CA Foundation Maths Solutions For Chapter 6 Sequence & Series

CA Foundation Maths Solutions For Chapter 6 Sequence And Series Introduction

Terms arranged in n definite order from ;i sequence. The terms in a sequence may lie numbers, letters, symbols or even words.

Here, we consider general sequences, before moving on to three specific mathematical sequences known as progressions i.e., arithmetic, geometric and harmonic before considering
number series.

CA Foundation Maths Solutions For Chapter 6 Sequences

A Sequence is a logically ordered list of elements related to each other by some relationship.

Identify the pattern followed by the terms in a sequence and use the pattern to find the terms of the sequence, sum of the terms in the sequence or to identify properties of the sequence.

Terms of a sequence are generally denoted by T₁,T₂, T₃,……. T_n

A lot of sequences either display a difference-based or a multiplicative pattern.

However, there are infinite ways in which sequences can be formed. Only a couple of patterns are discussed below to show how to analyze a sequence and mathematically represent it.

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Difference

In these sequences, consecutive terms are related to each other in terms of the difference between the two.

This difference can be constant or may follow a logical pattern itself.

For example, consider the sequence 1, 3,7,13, 21, 31

Observe that the difference between successive terms is 2,4,6,8, and 10.

Hence, the pattern followed by the difference is that they are all multiples of 2. Hence, the term after 31 should be 31+12=43. Mathematically, each term of the sequence can be expressed as:

T₁ =1+2×0=1
T₂=1+2×1=3
T₃ =3+2X2=7
T₄ =7+2X3=13
T₅ =13+2×4=21
T₆ =21+2X5=31

Thus, the n,h term of the sequence can be written as T_n = T_n 4+ 2(n — 1)

We can see that the nth term depends on the previous term as well as its position in the sequence.

Hence, T₇= 31+2×6=43

The advantage of representing a sequence mathematically is that later terms of the sequence can be found easily without the need to write the entire sequence.

For instance, in the above case, if T₃₈ is known and the value of T₄₀ is to be found, there is no need to write the entire sequence up to 40 terms. Instead, using the value of T₃₈, the value of T₃₉ and consequently, that of T₄₀ can be found in just 2 steps.

CA Foundation Maths Chapter 6 Sequence And Series Solutions

Solved Examples

Find the 7th term of the sequence 1,2,4,7,11, 16

Solution:

T₁ =1
T2=1+1=2
T3=2+2=4
T4=4+3=7
T5=7+4=11
T₆ =11+5=16

The nth term or this sequence can he expressed as

T_n =T_n-1+(n- 1)
T₇ =T₆ + (7- 1)
= 16+6
=22

Here the 7th term of this sequence would be 16+6=22

Cumulative Sequence

Consider the sequence 1,1,2,3,5,8,13, 21…. From the sequence observes that after the second term, the next term is the sum of the previous two terms. Hence the sequence mathematically represented in the following manner

T₁= 1
T₂=1
T₃ = 1+1=2
T₄ =2+1=3
T₅ =3+2=5
T₆ =5+3= 8
T₇ = 8+5=13
T₈ =13+8=21
and so on.

As can be seen, the next term is the sum of its previous two terms, hence

T_n = T_n-1 + T_n-2

∴T₉= T₈+ T₇

∴ T₉ =21+13=34

In these types of sequences, the pattern is formed with the help of its previous terms.

Solved Examples:

Find the next term of the series 3,4,11,24, 43

Solution:

Difference between the 1st and 2nd term=l

Difference between the 2ml and 3rd term=7=l+6

Difference between the 3rd and 4lh term=13=7+6

Difference between the 4th and 5lh term =19=13+6

Thus, the difference between the 5th and 6th terms=19+6=25.

Hence the next term is 43+25=68.

CA Foundation Maths Solutions For Chapter 6 Progressions

There are 3 specific types of sequences which show a specific mathematical relationship among their terms. These 3 types, also known as progressions, are named Arithmetic
Progression, Geometric Progression and Harmonic Progression

Arithmetic Progression

The terms of a sequence are said to be in Arithmetic Progression (A.P.) when they differ by a constant value known as their Common difference, dented by d. In other words, the
difference between any two consecutive terms in an A.P. is constant. The first term of an A.P. is generally denoted by a.

If d > 0, the A.P. is said to be an increasing A.P.

If d < 0, the A.P. is said to be a decreasing A.P.

If d = 0, each term of the sequence is equal.

Some A.P.s are given below.

1)1,2,3,4……where a= 1 and d= 1
2) 3.7, 11. 15…where a= 1 and d= 1
3) 8,2,-4,-10 … where a = 8 and d =-6

THE nth TERM OF AN A.P.

The first, second, third…nth terms of an A.P. are denoted by T₁, T₂ T₃,….T_n

For any arithmetic progression having the first term a and common difference d,

T₁ = a = a + [1-1) d
T₂ = a + d = a + (2-1) d
T₃ = a + 2d = a + (3-1) d and so on.

Continuing thus, the n,h term of an A.P. is, Tn=a+(n-1) d

Solved Examples

1. Find the fifteenth term of term A.P. -3,-9,-15,…….

Solution:

T_n=a+(n-1) d

Here, a = -3, d= -9-(-3) = -6 and n =15

T₁₅= -3 + (14)(-6)

T₁₅ = -87

2. The ninth term exceeds the fifth term of an A.P. by 32. The sum of the ninth and fifth terms is 114. Find the eighth term of the A.P.

Solution:

T₉ = a + (9 – 1)d = a + 8d

T₅ = a + (5- 1)d = a + 4d

T₉ + T₅ = 32

∴ (a+8d) – (a+4d)=32

∴ 4d=32

∴ d=8

T₉ + T₅ =(a +8d) + (a+4d) =2a+12d

T₉ + T₅ =114

114 = 2(a+6d)

a + 6d = 57

T8= a + 7d = a + 6d + d = 57 + 8 = 65

3. The 54th and the 4th terms of an A.P. are – 61 and 64 respectively. Find the 23rd term.

Solution:

The 54th and 4th term of the given A.P. can be represented as shown below.

a +53d= – 61 ……………………(i)

a + 3d = 64 ……………..(ii)

Subtracting we get

50d = -125

d= -5/2,

a= 64-3d=143/2

Hence the 23rd term = a+22d= 143/2+22(-5/2)=33/2

Sum of N Terms of An A.P.

Let the first term and common difference of an A.P. containing n terms he a and d respectively. Let Tn be the last term of the A.P. Then, the sum of n terms of the A.P. is
S_n= n/2[2a+ (n- 1)d] =n/2[a + T_n]

Solved Examples

1. The sixth and eighth terms of an A.P. are 38 and 52 respectively. Find the sum of the first twelve terms of the A.P.

Solution:

T₆ = a + 5d = 38

T₈ = a + 7d = 52

Solving the two equations, d= 7 and a=3

The sum of12 terms of the A.P. is

S₁₂=12/2[2×3 + (11)7] = 498

2. How many terms of the series -12, -9, -6, … must be taken so that the sum may be become 78?

Solution:

The given sequence is an A.P. with a = -12 and d =3

For an A.P. the sum of the first n terms is

s_n =n/2[2a + (n-l)d]

The sum ofn terms of this sequence is 78

n/2 x [2a + (n- 1)d] = 78

n/2 x [—24 + (n- 1)3] = 78

nx (n- 9) = 52

n=13

Average of the Terms of an A.P.

The average or the arithmetic mean ofn terms of an A.P

\(=\frac{S_n}{n}=\frac{n}{2} \times \frac{\left|a+T_n\right|}{n}\)

The average of n terms of an A.P

\(=\frac{a+T_n}{2}\)

The average of n terms of an A.P.

\(=\frac{(a+d)+\left(T_n-d\right)}{2}\)

The average of n terms of an A.P.

\(=\frac{(a+2 d)+\left(T_n-2 d\right)}{2}\)

Continuing thus, we see that the average of the terms of an A.P. is equal to the average of its first and nth terms, second and f(n-l),th terms, third and (n-2),th terms and so on.

In general, the average of the terms of the A.P. is equal to the average of the kth term from the beginning and the kth term from the end, or it is equal to average of any two terms
of the A.P. that are equidistant from the beginning and the end.

Also, if n is even the average of the A.P.is equal to the average of its

(n/2) th and (1+n/2)th terms.

if n is odd, the average of the terms of the A.P. is equal to the (n+1/2) th term of the A.P.

Solved Examples

1. The sum of the first nine terms of an A.P. is 387. Pind the fifth term,

Solution:

Since n is odd, the average of the first n terms of the A.P, is equal to the (n = 1)/2th

term = 5″‘ term.

Hence, the average of the first 9 terms of the A.P. = the (9 + 1)/2th term = 5,th term.

∴the 5′” term of the A.P. = 387/9= 43

If any two consecutive terms of an arithmetic progression are known, the series can be completely determined.

2. The fourteenth and fifteenth terms of an A.P. are 25 and 32 respectively. Find the 30th term, sum of the first 30 terms and the first term of the A.P.

Solution:

T₁₄ = a + 13d = 25

T₁₅ = a + 14d = 32

d = 7 and a = -s66

T30-66+29X 7 = 137

The sum of the first n terms = n/2 (a+T_n)

The sum of the first 30 terms

=30/2(-66+137)=1065

3. A teacher observes that the marks that the students in her class have scored are all different. She arranges her students in a line in increasing order of their marks such
that difference in marks scored by any two students next to each other is 4. The lowest marks that any student has scored are 11. The sum of the marks that all her students
have scored is 585. Find the marks scored by the student standing in the middle of the line.

Solution:

Since the difference in the marks of any two adjacent students is 4, the marks of the students standing in the line form an A.P. with a = 11 and d=4

Let there be n students in the class.

s_n = n/2 [2a + (n- l)d]

∴585 = n/2[22 + (n- 1)4]

∴(n) (2n+9) = 585

On solving this equation,

∴n = 15

The 8th student stands in the middle of the line. Since n = 15, the average of the first 15 terms is equal to the value of the (15+1)/2,h term = 8th term.

Hence, the value of the 8lh term of this A.P.

= 585/15 = 39

Hence, the marks of the 8lh student are 39.

Note:

When three terms are in Arithmetic progression, the middle term is the arithmetic mean of the other two. It is always convenient to take three terms in an A.P. as (a-d), a and (a + d).
Similarly, four terms in an A.P. could be taken as a-3d, a-d, a + d and a + 3d; five terms could be taken as a -2d, a – d, a, a+ d, a + 2d.

The advantage of representing terms in tin’s way is that the sum of terms is then obtained in only one unknown, i. e. when the sum of these terms is written mathematically, d cancels out and the sum is expressed only in terms of a.

Sequence And Series Chapter 6 Ca Foundation Answers

Some Properties of an A.P.

If each term of an A.P. is increased, decreased, multiplied or divided by the same non -zero number, then the resulting sequence is also an A.P. In case the terms are increased or decreased by some quantity, the common difference of the new A.P. remains equal to that of the original A.P.

In case the terms are multiplied or divided by a constant c (c≠0), the common difference d accordingly changes to d x c or d/c.

The number of elements in an arithmetic series from n₁ to n₂ with a step size (or common difference) ofm is1 = (n₁-n₂)/m

Solved Examples

1. How many multiples of 13 lie between 1000 and 5000? What is the sum of all these multiples?

Solution:

The lowest multiple of 13 that is greater than 1000 is 1001. The greatest multiple of 13 that is lesser than 5000 is 4992.

Here, m=13, n₁=1001 and n₂=4992

\(\text { There are } \frac{4992-1001}{13}+1\)

=308 multiples of 13 between 1000 and 5000. The sum ofall these multiples is

S=n/2(a+T_n)

S=308/2(1001+4992)

∴ S=9,22,922

If the sum of the first p terms of an A.P. is equal to the sum of the first q terms of the A.P. such that p and q are different, then the sum of (p + q) terms of the A.P. is zero.

2. The sum of the first 16 terms of an A.P. is equal to the sum of the first 24 terms of the A.P. Find the sum of the first 40 terms of the A.P.
Solution:

∴S₁₆=S24

∴S₍₁₆₊₂₄₎=0

∴S₄₀=0

CA Foundation Maths Solutions For Chapter 6 Geometric Progression

The terms of a sequence are said to be in Geometric progression (G.P.) when they increase or decrease by a constant factor. This constant factor is called the common ratio, denoted by r, and can be found by dividing any term of the sequence by the preceding term.

If the first term is positive and common ratio is greater than 1 (or if the first term is negative and the common ratio is less than1 and positive), the G.P. is an increasing G.P.

If the first term is positive and the common ratio is less than 1 and positive (or if the first term is negative and the common ratio is greater than 1), the G.P. is and decreasing G.P.

In other words, if all terms are greater than the preceding terms, the G.P. is an increasing G.P. else it is a decreasing G.P.

If the common ratio is equal to 1, all terms of the G.P. are equal.

A sequence with all terms equal is both, an A.P. and a G.P.

If the first term is a, the terms of the progression are a, ar, ar², ar³……………

THE nth TERM OF A.G.P

If T₁, T₂, T₃ ….Tn denote consecutive terms of a G.P. then

T₁= a= ar^1-1

T₂ – ar= ar^2-1

T₁= ar2= ar^3-1

Continuing thus, the n,h term of the geometric progression is given by T_n= ar^n-1

Solved Examples

1. Find the fifth tern if the G.P. whose first terms is 3 and the common ratio is 1/3.

Solution:

a=3 and r= (1/3)

The 5th term =ar^5-1= ar⁴=3(1/3)^5-1

=3/81

=1/27

2. The product of the first five terms of an A.P. is 28. Find the third term.

Solution:

ax arx ar²xar³xar⁴=28

a⁵r¹⁰ =28

(ar²)⁵= 28

ar² =⁵√28

∴the third term = ⁵√28

Alternatively, In such a case, assume the central term to be a and find the other terms from this point onwards, In this case, there are 5 terms in all. Hence, let the third term of this G.P. be a.

Hence, the 1st, 2nd 4th, and 5″‘ terms will be a/r², a/r, ar, and ar².

a/r²xa/rxaxarxar²=28

a⁵=28

The third term =a= ⁵√28

In case the G.P. has an even number of terms, one can take the two central terms to be a/r and ar and proceed from there.

Sum of N Terms of a G.P.

The sum of n terms or a G.P. with

r< 1 is a(1-rⁿ)/1-r The sum of n terms of a G.P. with

r <1 is a(rⁿ-1)/r-1 The sum of an infinite number of terms of a decreasing G.P.

=a/1-r

Geometric Mean

If terms

a₁,a₂,a_n are in G.P. then the Geometric mean G of these n terms is given by

G=ⁿ√a₁xa₂xa₃x…………..xa_n

If three terms are in G.P. then the middle term is the Geometric mean of the other two terms. If a, b and c are in G.P. (a, c > 0 or a, c < 0), then b is the geometric mean of a and c, and is given by b = ac or b² = ac

If n is even, the geometric mean of the terms of the G.P. is equal to the geometric mean of its (n/2) th and 1+(n/2)th terms.

If n is odd, the geometric mean of the terms of the G.P. is equal to the

(n+1/2)th term of the A.P.

Note:

While solving problems, three terms in G.P. can be assumed to be a/r, a and ar. Similarly, four terms in G.P. can be assumed to be a/r³, a/r, ar and ar³.

The advantage of representing terms in this way is that the product of terms is then obtained in only one unknown, i.e. in the product, r cancels out and the product is then expressed only in terms of a.

Some Properties of a G.P.

If each term of a G.P.is multiplied or divided by the some non-zero quantity, then the resulting sequence is also a G.P. with the common ratio remaining the

The reciprocals of the terms of a given G.P. also form a G.P. where the ratio is the reciprocal of that of the earlier G.P.

In a finite G.P. the product of two terms equidistant from the first and the last terms is same as the product of the first and the last term.

Exercise – 1: Arithmetic Progression (AP)

Choose the most appropriate option ( 1 ), ( 2 ), ( 3) or (4).

1. The nth element of the squence 1,3,5,7………………is

n
2n – 1
2n +1
none of these

Answer: (2) 2n – 1

Seq =1,3,5,7…

a = 1 d = 2

a_n = 1 + (n-l)d= 1 + [n-1)2

= 1 + 2n – 2 = 2n – 1

2. The nth element of the sequence -1, 2, -4, 8 is

(-1 )ⁿ2^n-1
2^n-1
2ⁿ
none of these

Answer: (1)(-1)ⁿ2^n-1

Seq = -1,2, -4, 8

a = 1

n = 1 (-1)1 21-1 =-l

n = 2 (-1)2 22-1 = 2

n = 3 (-1)3 23-1 = -4

a_n= (-1)ⁿ 2^n-1

3. \(\sum_{i=4}^7 \sqrt{2 i-1}\)

\(\sqrt{7}+\sqrt{9}+\sqrt{11}+\sqrt{13}\)
\(\sqrt[2]{7}+2 \sqrt{9}+2 \sqrt{11}+2 \sqrt{13}\)
\(\sqrt[2]{7}+2 \sqrt{9}+2 \sqrt{11}+2 \sqrt{13}\)
None of these

Answer: (1) \(\sqrt{7}+\sqrt{9}+\sqrt{11}+\sqrt{13}\)

\(=\mathrm{i}=4 \sum_{i=4}^7 \sqrt{8-1}=\sqrt{7}\) \(\sqrt{7}+\sqrt{9}+\sqrt{11}+\sqrt{13}\) \(=\mathrm{i}=5 \sqrt{2 i-1}=\sqrt{9}\)

4. The sum to 00 of the series -5,25,-125,625…..can be written are

\(\sum_{k=1}^{\infty}(-5)^k\)
\(\sum_{k=1}^{\infty} 5^k\)
\(\sum_{k=1}^{\infty}-5^k\)
none of these

Answer: (1)

\(\sum_{k=1}^{\infty}(-5)^k\)

= -5,25,-125,625

= (-5)¹, (-5)²,( 5)³

= (-5)^k

5. The first three terms of sequence when nth term tn is n2- 2n are

-1,0,3
1, 0, 2
-1,0,-3
none of these

Answer: (1) -1,0,3

= nth terms = tn = n² – 2n

=1st term = t₁ = 1 – 2 = -1

= 2nd term = t₂ = 4 – 4 = 0

= 3rd term = t₃ = 9 – 6 = 3

6. Which term of the progression -1,-3,-5,………Is -39

21st
20 th
19th
none of these

Answer:(2) 20 th

= a = -1 d = -2

a_n= -39

-1 + (n-1) (-2) = -39

(n-1) (-2) = -38

n – 1 = 19. Therefore, n = 20

7. The value of x such that 8x + 4, 6x- 2, 2x + 7 will form an AP is

15
2
15/2
none of the these

Answer: (3) 15/2

= d will be same if in AP

= d = a_n+1+an

= 6x – 2 – 8x- 4 = 2x + 7 – 6x + 2

-2x – 6 = -4x + 9

2x=15 Therefore, x =15/2

8. The mth term of an A. P. is n and nth term is m. The rth term ofit is

m + n +r
n + m- 2r
m + n + r/2
m + n – r

Answer: (4) m + n – r

1) = a_m = a+(m-1)d

a + dm- d = n

2) = a_n= a + (n-1)d

a + dn – d = m

Substracting 2) from 1]

d + dn – d = m

a + dm – d = n
– – +

a (n-m) = m-n

d = -1

a = m + n – 1

a_r = m + n -1 (r-1) (-1)

=m+n-1-r+1

= m + n – r

9. The number of the terms of the series………….will amount to 155 is

30
31
32
none of these

Answer: (1) 30 and (2) 31

\(=10+9 \frac{2}{3}+9 \frac{1}{3}+9 \ldots \ldots\) \(=10+\frac{29}{3}+\frac{28}{3}+9+\ldots \ldots\)

= a= 10 d= -1/3

S_n = n/2 (2a + (n- l)d)

155 =n/2(20 + (n-1)-1/3

310 = 20n -1/3n(n-1

310 = 20n-1/3(n2-n)

930 = 60n – n² + n

n² – 61n + 930 =

n(n-30) -31(n-30) = 0

n = 31 n = 30

10. The nth term of the series whose sum to n terms is 5n² + 2n is

3n – 10
10n – 2
10n – 3
none of these

Answer: (3)

= S_n = 5n² + 2n

S₁ = a₁ = 5+2 = 7

S₁ = a₁ + a₂ = 5(2)² + 2(2) = 24

S₂-S₁ = a₂

24-7 = a₂

a₂= 17

d = a₂-a₂= 10

a_n = 7 + (n-1) 10

= 7 + 10n – 10

= 10n – 3

11. The 20th term of the progression 1, 4, 7, 10,……is

58
52
50
none of these

Answer: (1) 58

= 1,4,7,10

a = 1, d = 3

a_n = 1 + (n-1)3

= 1 + 3n – 3

= 3n – 2

a₂₀ = 3(20)-2 = 58

12. The last term of the series 5, 7, 9,…..to 21 terms is

44
43
45
none of these

Answer: (3) 45.

= a₂₁ = 5 + (n-1)2

= 5 + (20)2

= 45

13. The last term of the A.P. 0.6, 1.2, 1.8,… to 13 terms is

8.7
7.8
7.7
none of these

Answer: (2) 7.8

= a₁₃ = 0.6 + (12) 0.6

= 0.6 (1+12)

= 0.6 x 13 = 7.8

14. The sum of the series 9, 5, 1,…. to 100 terms is

-18,900
18,900
19,900
none of these

Answer: (1) -18,900

S₁₀₀ = n/2 (2a + (n – 1)d) = 50(18 + 99(-4))

= 18-396

= -18,900.

15. The two arithmetic means between -6 ami 1 4 is

2/3,1/2
\(2 / 3,7 \frac{1}{3}\)
\(-2 / 3,-7 \frac{1}{3}\)
none of these

Answer: (2) 2/3, 7 1/3

Two A-M between – 6 & 14

a=-6 a4=14

a₄=-6+3d=14

3d=20 d=20/3

\(\mathrm{a}_2=-6+\frac{20}{3}=\frac{2}{3} \mathrm{a}_3=-6+\frac{40}{3}=+7 \frac{1}{3}\)

16. The sum of three integers in AP is 15 and their product is 80. The integers are

2, 8, 5
8, 2, 5
2, 5, 8
8, 5, 2

Answer: (3) & (4)

= let 3 integers be a – d, a, a + d

Sum = a-d + a + a + d = 3a=15->a = 5.

(a – d) (a + d) a = 80

(a² – d²)a = 80

a² – d² = 16

a² – 16 = d²

25 – 16 = d² = 9 -7 d =±3.

AP when a = 5, d = 3

2,5,8

AP when a = 5, d = -3

8,5,2.

17. The sum of n terms of an AP is 3n² + 5n. The series is

8,14,20,26
8, 22, 42, 68
22, 68, 114,
none of these

Answer: (1)

S₁ = 3 + 5 = 8 = a,

S₂ = 12 + 10 = 22

a₂ = S2 – S1 = 14, d = 6

Series -> 8, 14, 20,……..

18. The number of numbers between 74 and 25,556 divisible by 5 is

5,090
5,097
5,095
none of these

Answer: (2) 5097

-7 75.80,……….25,555

It forms an AP with a = 75 d = 5

25,555 = Last term = a_n

A_n = 75 + (n-1)5

25,555 = 75 +(n-1)5

25480 = (n-1)5

5096 = n-1

5097 = n.

19. The pth term of an AP is (3p- 1)/6. The sum of the first n terms of the AP is

n(3n + 1)
n/12 (3n + 1)
n/12 (3n – 1)
none of these

Answer: (2) n/2(3n + l)

\(\rightarrow \mathrm{a}_1=\frac{3(2)-1}{6}=\frac{2}{6}=\frac{1}{3}\) \(a_2=\frac{3(2)-1}{6}=\frac{5}{6}\) \(d=\frac{5}{6} \frac{-2}{6}=\frac{3}{6}=\frac{1}{2}\) \(3_n=\frac{n}{2}(2 a+(n-1) d)=\frac{n}{2}\left(\frac{2}{3}+\right.\) \(\left.(n-1) \frac{1}{2}\right) \text {. }\) \(\frac{n}{2}\left(\frac{2}{3}+\frac{1}{2} n-\frac{1}{2}\right)=\frac{n}{2}\left(\frac{1}{6}+\frac{1}{2} n\right)\) \(=\frac{n}{12}(1+3 n)\)

20. The arithmetic mean between 33 and 77 is

50
45
55
none of these

Answer: (3) 55.

AM = a + C

\(A M=\frac{a+c}{2}=\frac{33+77}{22}=\frac{110}{2}=55\)

21. The 4 arithmetic means between -2 and 23 are

3, 13, 8, 18
18, 3, 8, 13
3, 8, 13, 18
none of these

Answer: (3) 3,8,13,18

->-2,_,_,_,_,_,23

a = 2

a₆= 23 = a + 5d = 23

-2 + 5d = 23

d = 5.

a₂=-2+5=3

a₃=3+5=8

a₄=8+5=13,

a₅=13+5=18.

Given = -2,3,8,13,18,23.

22. The first term of an A.P is 14 and the sums of the first five terms and the first ten terms are equal is magnitude but opposite in sign. The 3rd term of the AP is

6×4/11
6
4/11
none of these

Answer: \(6 \frac{4}{11}.\)

-> S₅ = -S₁₀

\(\frac{5}{2}(28+4 d)=-\left(\frac{10}{2}\right)(28+9 d)\)

28 + 4d = -56 -18d

22d = -84

d=-84/22=42/11

a₃=14+2(-42/11)

=70/11

= 6×4/11

23. The sum of a certain number of terms of an AP series -8, -6, -4,……is 52. The number of terms is

12
13
11
none of these

Answer: (2) 13.

S_n = 52

a = -8, d==2

S_n =n/2 (-16 + (n-1)2)

52 x 2 = n(-16+2n-2) = 104

= -18n + 2n²

2n²- 18n-104 = 0

n²- 9n – 52 = 0

n²- 13n + 4n – 52 = 0

n(n-13)+4(n-13)=0

n = 13

24. The first and the last term of an AP are -4 and 146. The sum of the terms is 7171. The number of terms is

101
100
99
none of these

Answer: (1) 101

S_n = n/2(a + 1)

7171=n/2(-4 + 146)

14342 = n(142)

n = 101

25. The sum of the series 3 x1/2 + 7 + 10 x1/2 + 14 + …. to 17 terms is

530
535
535×1/2
none of these

Answer: (3) 535×1/2

a = 3.5 d=3.5

S₁₇ = 17/2(3.5×2 + (17 -1)3.5)

= 17/2(7 + 56) = 535×1/2

26. Three numbers are in AP and their sum is 21. If 1, 5, 15 are added to them respectively, they form a G. P. The numbers are

5, 7,9
9,5,7
7, 5,9
none of these

Answer: (1) 5,7,9

= S = 3a = 24 = a = 7

If 1,5,15 are added terms are

a-d + 1, a + 5, a + d + 15

8 – d, 12, 22+d -> GP

∴12/8-d=22+d/12

144 = (22+d) (8-d)

144=176-22d+8d-d²

d² + 14d – 32 = 0

d² + 16d – 2d – 32 = 0

d(d+16)-2(d+16) =0

d = 2, d = -16

If a = 7, d = 2

= AP = 5,7,9

If a= 7, d= -16

AP = 23,7,-9

27. If p, q and r are in A.P. and x, y, z are in G.P. then xq-r. yr-q. Zp-q is equal to

0
-1
1
none of these

Answer: (3) 1

q-p = r-q

p-q = q-r

2q = p+r

y² = xz

x^q-r.y^r-p.z^p-q

x^p-q.z^p-q.y^r-p

(x·z)^p-q.y^r-p

(y)^2p-2q.y^r-p

y^2p-2Q+r-P

y^p+r-2q→y^p+r-2q →y^2q-2q→y⁰= 1 (p +r = 2q)

28. The sum of 3 numbers in A.P. is 15. If 1, 4 and 19 be added to them respectively, the results are is G. P. The numbers are

-26, 5, -16
2, 5, 8
5,8,2
none of these

Answer: (1),(2)

a-d, a, a+d

be in AP

a-d+a+d+a=15

a=5

if 14,19 added

terms 6-d, 9, 24+d = GP

81 = (6-d)(24+d)

81 = 144 + 6d – 24d – d²

d² + 18d – 63 = 0

d² + 21d -3d – 63 = 0

d(d+21) -3 (d+21) = 0

d=21,d=3, -> d=3, and d=-21

Series = 2,5,8 if a = 5, d=3

Series = -26,5,-16 if a = 5, d = -21

29. If the terms 2x, (x+10) and (3x+2) be in A.P., the value of x is

7
10
6
none of these

Answer: (3) 6.

same common difference = d

x + 10 – 2x = 3x + 2 – x- 10

10-x = 2x-8

18 = 3x

x = 6

30. If A be the A.M. of two positive unequal quantities x and y and G be their G. M, then

A < G
A>G
A ≥ G
A ≤ G

Answer: (b) A > G.

If two unequal quantities present

AM > GM

A > G

CA Foundation Maths Chapter 6 Detailed Solutions

31 The A.M. of two positive numbers is 40 and their G. M. is 24. The numbers are

(72, 8)
(70,10)
(60,20)
none of these

Answer: (1) 72,8

AM=a+c/2 ac=GM²

80 = a + c ac = 576

\(80=\frac{576}{c}+c \quad a=\frac{576}{c}\)

80c = 576 + c²

C²- 80c + 576 = 0

C²- 72c- 8c + 576 = 0

C(c-72)-8 (c-72) = 0

C = 72, C = 8

If C = 72, a = 8

If C = 8, a= 72,8

= 72,8

32. Three numbers are in A.P. and their sum is 15. If 8, 6, 4 be added to them respectively, the

2, 6, 7
4,6,5
3,5,7
none of these

Answer: (3) 3,5,7

Sum = 15

13a = 15s = a= 5

If 8,6,4 is added respectively

13-d, 11,9 + d -> GP

121 = (13-d) (9+d)

121 = 117 + 13d -9d -d²

d² – 4d + 4 = 0

d²- 2d -2d + 4 = 0

d(d-2)-2(d-2) = 0

d = 2.

Series -> 3,5,7.

33. A sum of is paid off in 30 instalments such that each instalment is *?10 more than the proceeding instalment. The value of the 1st instalment is

36
30
60
none of these

Answer: (4) none of the above

S₃₀ = 6240

6240 =30/2(2a + 290)

416 = 2a + 290

2a =126 a = 63

34. If x, y, z are in A.P. and x, y, (z + 1) are in G.P. then

(x- z)² = 4x
z² = (x- y]
z = x- y
none of these

Answer: (1)

x,y, (z+1) = GP

y² = (z + 1)x

2y = x +z

y = x + z

\(y^2=\left(\frac{x+2}{4}\right)^2 \rightarrow 4(z+1) x=(x+z)^2\)

4xz + 4x = x² + z²+ 2zx

2xz + 4x = x² + z²

4x = x² + z² – 2xz

(x- z)²= 4x

35. The numbers x, 8, y are in G.P. and the numbers x, y, -8 are in A.P. The value of x and y are

(-8, -8)
(16,4)
(8,8)
none of these

Answer: (1) (2)

X,8,y = GP

X,y,-8 = AP

2y-64/2 = —8 -> 2y²- 64 = -8y

2y² + 8y- 64 = 0

Y² + 4y- 32 = 0

Y² + 8y- 4y- 32 = 0

y(y+8) – 4y- 32 = 0

y(y+8) – 4 (y+8) = 0

y = 4 y=-8

x= 16 x=-8

36. The sum of all odd numbers between 200 and 300 is

11,600
12,490
12,500
24,750

Answer: (4) 12,500

201,203,…….299

AP with a = 201, d = 2

I = 299, a = 201, d = 2

299 = 201 + (n – 1) 2

98= (n- 1)2

49 = n-1= n = 50

S₅₀= n/2 (a + 1) = 25 (201 + 299)

= 12,500

37. The sum of all natural numbers between 500 and 1000 which are divisible by 13, is

28,405
24,805
28,540
none of these

Answer: (1) 28,405

= 507,520,……. 988

AP, a = 507 d = 13 I = 988

988 = 507 + (n – 1) 13

481 = (n – 1) 13

37 = n-1 = n = 38

S₃₈ = 38/2 (507 + 988) = 28,405

38. If unity is added to the sum of any number of terms of the A.P. 3, 5, 7, 9…….the resulting sum is

‘a’ perfect cube
‘a’ perfect square
‘a’ number
none of these

Answer: (2) Perfect square

If 1 is added to any sum

It is same as adding1 as

A term of AP, So AP becomes

AP: 1,3, 5, 7……….

These are n natural odd number

Now sum of first n natural odd no. is n2 i.e. perfect square

39. The sum of all natural numbers from 100 to 300 which are exactly divisible by 4 or 5 is

10,200
15,200
16,200
none of these

Answer: (3) 16,200

Natural no. divisible by 4 or 5

= natural no divisible by 4 + natural no divisible by 5

= natural no divisible by 4 & 5 i.e. 20

100, 104………..300

AP -> a = 100 d = 41 = 300

N = 51 terms

= S₅₁ = 51/2 (100+300)

Sum of natural no divisible by 4 = 10200

For divisible by 5

100,105,…..300

AP, a = 100 I = 300 d = 5 n=41

S₄₁ = 41/2 (100 + 300) = 8200

Sum of natural no divisible by 5 = 8200

For4&5

100,120,…..300

AP a = 100 d = 20, I = 300, n = 11

S₁₁ = 11/2 (100 + 300) = 2200

∴ sum of natural no divisible by 4 or 5

= 10200 + 8200-2200= 16200

40. The sum of all natural numbers from 100 to 300 which exactly divisible by 4 and 5 is

2,200
2,000
2,220
none of these

Answer: (1) 2200

Divisible by 4 & 5

i.e. divisible by 4 x 5 = 20

100,120……300

AP -> a = 100 d = 20 I = 300 n = 11

S₁₁ =11/2 (100 + 300)

= 2200

41. A person pays 975 by monthly instalment each less then the former by 5. the first instalment is 100.The time by which the entire amountwill be paid is

10 months
15 months
14 months
none of these

Answer: (2) 15 months

a = 100

d = -5

100,95,…….

S_n = 975, n =?

975 =n/2 (200 + (n – 1) (-5))

1950 = n (200 – 5n + 5)

1950 = 205 – 5n²

5n²- 205n+ 1950 = 0

n² – 41n + 390 = 0

n²-26n- 15n + 390 = 0

n (n – 26) – 15 (n-26) = 0

n = 15 or n = 26

15 months

42. A person saved 16,500 in ten years. In each year after the first year he saved 100 more than he did in the precedinh year. The amount of money he saved in the 1st year was

1000
1500
1200
none of these

Answer: (3)

d = 100

S₁₀ = 16500

16500 = 10/2 (2a + 9 (100))

33000 = 10 (2a + 900)

3300 -900 = 2a

a = 1200

CA Foundation Maths Solutions For Chapter 6 Exercise – 2: Geometric Mean

Choose the most appropriate option (1), (2), (3) or (4)

1. The 7th term of the series 6, 12, 24,……… is

384
834
438
none of these

Answer: (1) 384

GP: 6, 12, 24,

A = 6 r=12/6=2

a₇ = ar^7-1 = ar⁶

= 6(2)⁶ =384

2. t₈ of the series 6, 12, 24,…is

786
768
867
none of these

Answer: (2) 768

GP: 6, 12, 24….

a = 6, r = 2

a₈= ar^8-1 = ar⁷

= 6 (2)⁷ =768

3. t₁₂ of the series -128, 64, -32, ….is

– 1/16
16
1/16
none of these

Answer: (3) 1/16

t₁₂ = a¹¹ =- 128 (-1/2 )11 =1/16

4. The 4th term of the series 0.04, 0.2, 1, … is

0.5
1/2
5
none of these

Answer: (3) 5

0.04, 0.2, 1,………….

a = 0.04 r= 5

a₄=ar³

=0.04(5)³ = 5

5. The last term of the series 1, 2, 4,…. to 10 terms is

512
256
1024
none of these

Answer: (1) 512

a = 1, r=2

a₁₀ = ar⁹ = 1(2)⁹ = 512

6. The last term of the series 1, -3, 9, -27 up to 7 terms is

297
729
927
none of these

Answer: (1) 729
a = 1, r = -3

a₇ = ar⁶= 1(-3)⁹ =729

7. The last term of the series x2, x, 1, …. to 31 terms is

x²⁸
1/x
1/x²⁸
none of these

Answer: (3) 1/x²⁸

a = x², r = 1/x

a₃₁=ar₃₀=x²(1/x)³⁰

=x²x1/x³⁰=x^-28=1/x²⁸

8. The sum of the series -2, 6, -18, …. to 7 terms is

-1094
1094
– 1049
none of these

Answer: (1) -1094

a = -2, r = -3

a₇= ar⁶ = -2(-3)⁶ = -1458

GP = -2, 6,-18,……-1458

S_n =Ir-1/r-1 =-1094

9. The sum of the series 243, 81, 27, …. to 8 terms is

36
\(\left(36 \frac{13}{30}\right)\)
\(36 \frac{1}{9}\)
none of these

Answer: (4) none of these

a = 243 r = -1/3

\(S_8=\frac{a\left(1-r^n\right)}{1-r}=\frac{243\left(1-\frac{1}{3}_3^8\right)}{1-\frac{1}{3}}\) \(=\frac{243(0.999847) .3}{2}\)

= 364.4445 = 364.45

10. The sum of the series \(\frac{1}{\sqrt{3}}+1+\frac{3}{\sqrt{3}}+\ldots . . \text { to } 18 \text { terms is }\)

\(9841 \frac{(1+\sqrt{3})}{\sqrt{3}}\)
9841
\(\frac{9841}{\sqrt{3}}\)
none of these

Answer:

\(9841 \frac{(1+\sqrt{3})}{\sqrt{3}}\) \(a=\frac{1}{\sqrt{3}}, \quad r=\sqrt{3}\) \(S_{18}=\frac{a\left(r^n-1\right)}{r-1}=\frac{1}{\sqrt{3}} \frac{\left(\left(\sqrt{3)}^{18}-1\right)\right.}{\sqrt{3}-1}\) \(\frac{1}{\sqrt{3}} \frac{\left(\sqrt{3}^{18}-1\right)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{1}{\sqrt{3}} \frac{\left(\sqrt{3}^{18}-1\right)(\sqrt{3}+1)}{3-1}\) \(=\frac{1}{2 \sqrt{3}}\left(\sqrt{3}^{18}-1\right)(\sqrt{3}+1)\) \(=9841 \frac{(1+\sqrt{3})}{\sqrt{3}}\)

11. The second term of a G P is 24 and the fifth term is 81. The series is

16,36,24, 54,..
16, 24,36,54,..
24,36, 53,…
none of these

Answer: (3)

16, 24, 36, 54

ar = 24 ar⁴ = 81

\(\frac{a r^4}{a r}=\frac{81}{27}=r^3=\frac{81}{27}\) \(=r^3=\left(\frac{3}{2}\right)^3 \quad r=\frac{3}{2}\) \(\mathrm{ar}=24=\mathrm{a}=\frac{24.2}{3}=16\)

12. The sum of 3 numbers of a G P is 39 and their product is 729. The numbers are

3, 27, 9
9,3, 27
3, 9. 27
none of these

Answer: (3)

Sum = 39 product =729

a/r, a, ar

a/r.a.ar=729

a³=729

a/r+a+ar=39

9/r+9+9r=39

9(1/r+1+r)=39

\(9\left(\frac{1+r+r^2}{r}\right)=39\)

9 (1 + r + r²) = 39x

9 + 9x + 9x² = 39r

9r²-30r + 9 = 0

9r²- 27r- 3r + 9 = 0

9r(r-3)-3 (r-3) = 0

r=3/9 r=3

when, r = 3, GP = 3, 9,27

When r= 1/3 GP = 27, 9,3

13. In a G. P, the product of the first three terms 27/8. The middle term is

3/2
2/3
2/5
none of these

Answer: (1) 3/2

GP= a/r, a, ar

a/r.a.ar=27/8=a³=27/8=a=3/2

14. If you save 1 paise today, 2 paise the next day 4 paise the succeeding day and so on, then your total savings in two weeks will be

163
183
163.83
none of these

Answer:(3) 163.83 Rs

a₁₄ = ar¹³, a = 1, r = 2

= 1(2)¹³ =8192

\(S_{14}=\frac{I r-a}{r-1}=\frac{16384-1}{1}=16383 \text { paise }\)

= 163.83 Rs.

15. Sum of n terms of the series 4 + 44 + 444 + …………. is

4/9 {10/9 (10ⁿ -1) -n}
10/9 ( 10ⁿ -1 ) -n
4/9(10ⁿ-!) -n
none of these

Answer:(1) 4/9 {10/9 (10ⁿ -1) -n}

4 + 44 + 444 +………

4/9(9 + 99 + 999 +…..)

=4/9[(10-1)+(10²-1)+(10³-1)+…..(10ⁿ-1)]

=4/9(10+10²+10³……10ⁿ-n)

=4/9(10+10²+10³……10ⁿ-1)-n)

\(=\frac{4}{9}\left(10\left(\frac{10^n-1}{9}\right)-n\right)\) \(=\frac{4}{9}\left(\frac{10}{9}\left(10^n-1\right)-n\right)\)

16. Sum of n terms of the series 0.1 + 0.11 + 0.111 + … is

1/9 {n – ( 1-(0.1)ⁿ)}
1/9 {n- (1-(0.1)ⁿ)/9)
n-1 – (0.1)ⁿ/9
none of these

Answer:(2) 1/9 {n- (l-(0.1)ⁿ)/9)

= 0.1 +0.11 + 0.111 +……….

= 1/9(0.9 + 0.99 +……)

= 1/9[(1-1/10)+(1-1/10²+…..)

= 1/9[n-(1/10+1/10²+……)]

= 1/9[n-1/10(1+1/10+1/10²+…….)]

\(=\frac{1}{9}\left[n-\frac{1}{10}\left(\frac{1-\frac{1}{10}}{1-\frac{1}{10}}\right)\right]\) \(=\frac{1}{9}\left[n-\frac{1}{10}\left(\frac{1-10^{-n}}{9}\right) 10\right]\) \(=\frac{1}{9}\left[n-\left(\frac{1-10^{-n}}{9}\right)\right]\)

=1/9 {n- (l-(0.1)ⁿ)/9)

17. The sum of the first 20 terms of a G. P is 244 times the sum of its first 10 terms. The common ratio is

±√3
±3
√3
none of these

Answer:(1) ±√3

\(\frac{a\left(r^{20}-1\right)}{r-1}=244 \frac{(a)\left(r^{10}-1\right)}{r-1}\)

r²⁰ – 1 = 244 (r¹⁰– 1)

y² – 1 = 244 (y- 1)

y² – 1 = 244y – 244

y² – 244y + 243 = 0

y² – 243y – y + 243 = 0

y (y – 243) – 1 (y – 243) = 0

y = 243 y=1

y = r¹⁰

r¹⁰ = 243 r¹⁰ = 1

(±√3)¹⁰=243 r=±1

r= ±√3

18. Sum of the series 1 + 3 + 9 + 27 +….is 364. The number of terms is

5
6
11
none of these

Answer: (2) 6

\(S n=\frac{a\left(r^n-1\right)}{r-1}\) \(=364=\frac{1\left(3^n-1\right)}{2}\)

19. The product of 3 numbers in G P is 729 and the sum of squares is 819. The numbers are

9, 3, 27
27, 3, 9
3, 9, 27
none of these

Answer: (3) 3, 9, 27

Product 729 ; sum of squares = 819

a/r. a . ar = 729

a³ = 729

a = 9

a/r,9,9r = 9²/r² + 81 + 81r² = 819

81(1/r²+1+ r²) = 819

81 (1 + r² + r⁴) = 819r²

81 + 81r- + 81r⁴ = 819r²

81r⁴ – 819r² + 81r² + 81 = 0 = 81 r⁴

738r² + 81 = 0

9r⁴ – 82r²+ 9 = 0

9r⁴ – 81r² — 1 r² + 9 = 0

\(9 r^2\left(r^2-9\right)-1\left(r^2-9\right)=r^2 \frac{1}{9}, r^2=9=r \frac{ \pm 1}{3}\)

r = ±3

When ± 3, a = 9

GP = 3,9,27 & -3,9,-27

When ± 1/3 a=9

GP= 27,9,3 & -27,9,-3

20. The sum of the series 1 + 2 + 4 + 8 + .. ton term

2ⁿ-1
2n – 1
1/2ⁿ -1
none of these

Answer: 2ⁿ – 1

\(S_{11}=\frac{1\left(2^n-1\right)}{2-i}=2^n-1\)

21. The sum of the infinite GP 14, – 2, + 2/7,- 2/49, + … is

\(4 \frac{1}{12}\)
\(12 \frac{1}{4}\)
12
none of these

Answer: \(12 \frac{1}{4}\)

\(S_n=\frac{a}{1-r}=\frac{14}{1+\frac{1}{7}}=\frac{7 \times 14}{8}=\frac{49}{4}=12 \frac{1}{4}\)

22. The sum of the infinite G. P. 1 – 1/3 + 1/9 – 1/27 +… is

0.33
0.57
0.75
none of these

Answer: (3) 0.75

\(\mathrm{S}_{\mathrm{n}}=\frac{a}{1-r}=\frac{1}{1+\frac{1}{3}}=\frac{3}{4}=0.75\)

23. The number of terms to be taken so that 1 + 2 + 4 + 8 + will be 8191 is

10
13
12
none of these

Answer: (2) 13

s_n= 8191

a = 1 r = 2

\(S_n=\frac{1\left(2^n-1\right)}{1}=8191\)

2ⁿ = 8192 n=B

24. Four geometric means between 4 and 972 are

12,36, 108,324
10,36, 108,320
12, 24,108,320
none of these

Answer: (1)

a = 4 I = 972

ar^n-1 = 972

ar⁵ = 972

4r⁵ = 972

r⁵ = 243 = r = 3

4,12,36,108,324, 972

25. The sum of 1 + 1/3 + 1/32 + 1/33 + … + 1/3n-1 is

2/3
3/2
4/5
none of these

Answer: (b)3/2

a = 1 x =1/3

\(S=\frac{a}{1-x}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}\)

26. The sum of the infinite series 1 + 2/3 + 4/9 + .. is

1/3
3
2/3
none of these

Answer: (2) 3.

a = 1 x = 2/3

\(S=\frac{a}{1-x}=\frac{1}{1-\frac{2}{3}}=\frac{3}{1}=3\)

27. The sum of the first two terms of a G.P. is 5/3 and the sum to infinity of the series is 3. The common ratio is

1/3
2/3
– 2/3
none of these

Answer: (2) & (3)

\(\frac{a}{1-x}=3\)

a = 3(l-x) = 3 -3x

a(l+x)=5/3

(3-3X)(1+X)=5/3

3(1-X)(1+x)=5/3==1-X²=5/9

\(1-\frac{5}{9}=x^2=\frac{4}{9}=x^2=x=\frac{2}{3}\)

How To Solve Sequence And Series In CA Foundation

CA Foundation Maths Solutions For Chapter 6

28. The sum of three numbers in G.P. is 70. If the two extremes by multiplied each by 4 and the mean by 5, the products are in AP. The numbers are

12, 18, 40
10, 20, 40
40, 20, 10
none of these

Answer: (2) & (3)

a + ax + ax² = 70

A.P = 4a, 5ax, 4ax²

10ax = 4a + 4ax²

10ax = 4a(1+x²)

5/2 x=1+x²

2x²- 5x + 2 = 0

2x²- 4a- x + 2 = 0

2x(x- 2) — 1(x — 2) = 0

x= 1/2,X = 2.

29. Given x,y,z. are in G.P. and Xp = Yq = Zσ then 1/p,1/q.1/a are In

A.P
G.P
Both A.P & G.P
None of these

Answer: (1)AP

X^p = Y^q = Z^σ

y² = xz

X^p = Y^q = Z^σ = k

\(x=k^{\frac{1}{p}}, y=k^{\frac{1}{q}}, z=k^{\frac{1}{\sigma}}\) \(k^{\frac{2}{q}}=k^{\frac{1}{p}} \cdot k^\sigma\) \(k^{\frac{2}{q}}=k^{\frac{1}{p}}+\sigma\) \(\frac{2}{9}=\frac{1}{p}+\frac{1}{\sigma}=2\left(\frac{1}{q}\right)=\frac{1}{p}+\frac{1}{\sigma}\) \(\frac{1}{p}, \frac{1}{q}, \frac{1}{\sigma} \text { are in AP. }\)

30. The sum of 1.03 + ( 1.03 )²+ ( 1.03 )³+ …. to n terms is

103 {(1.03)ⁿ– 1}
103/3 {(1.03)ⁿ– 1}
(1.03)n -1
none of these

Answer: (2)

1.03(1 + (1.03) +(1.03)² +……(1.03)ⁿ-1)

\(1.03\left(\frac{(1.03)^n-1}{1.03-1}\right)\) \(\frac{1.03}{0.03}\left((1.03)^n-1\right)\) \(\left.=\frac{103}{3}\right)\left((1.03)^n-1\right)\)

31. The nth term of the series 16, 8, 4…..is 1/217. The value of n is

20
21
22
none of these

Answer: (3) 22

a = 16 r = 1/2.GP

ar^n-1=a_n=1/2¹⁷

\(\frac{16}{2^{n-1}}=\frac{1}{2^{17}}=\frac{1}{2^{n-1}}=\frac{1}{2^{17} \cdot 2^4}\) \(=\frac{1}{2^{n-1}}=\frac{1}{2^{21}}\)

n- 1 = 21 -> n = 22

32. The sum of n terms of a G.P. whose first terms 1 and the common ratio is 1/2 , is equal to 1×127/128. The value of n is

7
8
6
none of these

Answer: (2)8

\(S_{n}=1 \frac{127}{128}=\frac{255}{128}\) \(\frac{\left(1-0.5^n\right)}{1-0.5}=\frac{255}{128}\) \(\frac{\left(1-0.5^n\right)}{0.5}=\frac{255}{128}\) \(1-0.5^n=\frac{127.5}{128}\)

33. t₄ of a G.P. in x, t₁₀ = y and t₁₆ = z. Then

x² = yz
z² = xy
y² = zx
none of these

Answer: (3)

T₄ = X, t₁₀ = y,t₁₆ = z.

ax³ = x ax⁹ = y ax¹⁵ = z

y² = a²x¹⁹

y² = xz

34. 1 f x, y, z are in G.P., then

y²= xz
y (z² + x²) = x(z² + y²)
2y = x+z
none of these

Answer: (1)

x,y,z in GP

y² = xz

35. At 10% C.I. p.a., a sum of money accumulate to 9625 in 5 years. The sum invested initially is

5976.37
5970
5975
5370.96

Answer: (1)

Sum of money at end 9625

Time series

Initial investment =? = a

Amt = a

Amount at end first yr= a + 0.1a = a(1.1)

Amount at end second yr = a(1.1) + a(0.1)(1.1)

= a(1.1+0.11)

= a(1.21) = a(1.1)²

Amount at end of third = a(1.1)³

Amount at end of n = a(1.1)ⁿ

9625 = a(1.1)ⁿ

9625 = a(1.1)⁵

a = 5976.37

36. The population of a country was 55 crores in 2005 and is growing at 2% p.a C.I. the population is the year 2015 is estimated as

5705
6005
6700
none of these

Answer:

(3) 6700

Population initial = 55

Time 10 years

Population at end ofyr= 55 + 55(0.2) = 55(1.02)

At end ofyr 2 = 55(1.02) + 55(1.02)

(0.02) = 55(1.02)²

At end ofyrs = 55(1.02)ⁿ

S₁₀ = a(l + r)ⁿ = 55(1.02)¹⁰

S₁₀ = 6700.

CA Foundation Maths Solutions For Chapter 6 Exercise – 3 (Mix- AP and GP)

1. What is the arithmetic mean of the arithmetic progression 6,8,10,12,14,16?

Answer: Arithmetic mean = sum of the terms/number of terms

\(=\frac{6+8+10+12+14+16}{6}\)

= 66/6

= 11

Choice (2)

CA Foundation Maths Solutions For Chapter 6

2. What is the geometric mean of the geometric progression 2,4,8,16?

32
√32
64
8

Answer:

Geometric mean= [(2)(4)(8)(16)]^1/4

=[(21)(22)(23)(24)]^1/4=2^10/4 = √32

Alternative solution:

Geometric mean of a geometric progression

\(=\sqrt{\text { first term } \times \text { last term }}\) \(=\sqrt{(2)(16)}\)

=32

Choice (2)

3. What is the fourth term of the arithmetic progression in which the first term is 4 and the seventh term is 28?

Answer: The fourth term is equidistant (3 teams away) from the first and the seventh term in the arithmetic progression.

the fourth term is the arithmetic mean of the first term and the seventh term in an A.P.

fourth term \(=\frac{4+28}{2}\)

32/2=16

choice (1)

4. What is the sum to 7 terms of the arithmetic progression in which the first term is 2 and the common difference is 4?

Answer: Sum of the first n terms of an A.P

\(=s_n=\left(\frac{n}{2}\right)[2 a+(n-1) d]\) \(s_n=\left(\frac{7}{2}\right)[2(2)+(7-1) 4]\) \(=\left(\frac{7}{2}\right)[4+24]=\frac{7}{2}(28)=98\)

choice (2)

5. What is the sum to 8j terms of an arithmetic progression in which the first term is 3 and the last term is 31?

136
58.5
132
Cannot be determined

Answer: Given first term=3 and the last term=31

As we don’t know the total number of terms, we cannot find the common difference.

we cannot find the sum of first 8 terms of the Progression, choice (4)

6. What is the arithmetic mean of an arithmetic progression with 13 terms, in which the 7lh term is 9?

9
91/7
95/7
Cannot be determined

Answer: Given total number of terms is 13, i.e. n=13 in an A.P. if n is odd

then \(\frac{(n+1)^{t h}}{2}\) term is the arithmetic mean of that A.P.

as \(\frac{(13+1)}{2}=7^{t h}\) ,term is the arithmetic mean

Given 7th term is 9

9 is the arithmetic mean

Choice (1)

7. What is the sum to 15 terms of an arithmetic progression’ whose 8th term is 4?

30
60
40
Cannot be determined

Answer: The 8th term is equidistant from first term and 15 terms

8th term is the arithmetic of first 15 terms sum of first 15 terms = 15x (eight term)=15(4)

=60

Choice (2).

8. What is the sum of all the terms in an arithmetic progression in which the first term is 5, the last term is 15 and the number of terms is 11?

55
110
115
Cannot be determined

Answer: in A.P

S_n=(n/2)[first term+last term]

S_n=(11/2)[5+15]=110

Choice (2)

9. What is the seventh term of an arithmetic progression whose first term is 9 and the common difference is 3?

Answer: in an A.P. n 1,1 term (t_n) = a+ (n-1) d

t₇ = 9 + (7- 1)3 = 9 + 6(3) = 27

choice (1)

10. What is the fourth term so a ‘geometric progression in which the second term is 4 and the sixth term is 64?

Answer: fourth term is equidistant from the second term and the sixth term.

in a G.P. the fourth term is the geometric mean of the second term and the sixth terms

Fourth term = \(\sqrt{(\text { second term)(sixth term) }}=\sqrt{(4)(64)}=16\)

choice (4).

11. What is the sum to 4 terms of a geometric progression whose first term is 6 and the common ratio is 3?

Answer: In a G.P.

\(S_n=\frac{a\left(1-r^n\right)}{1-r}\) where is the common ration, a the first term and sn in the sum to n term.

\(S_4=\frac{6\left(1-3^4\right)}{1-3}=\frac{6(1-81)}{2}\)

=(3)(80)= 240

Choice (4)

12. In an arithmetic progression having 100 terms, the m th term form the beginning and the mthTerm from the end is 10 and 20 respectively what is the sum of all the terms?

3000
1500
3200
Cannot be determined

Answer:

As the average of the kth term from the beginning and the kth term from the end is equal to the arithmetic mean the average of 10 and 20 is the arithmetic mean i.e. the mean is

\(\frac{10+20}{2}=15\)

in an A.P. as S_n = n (AM.)

S₁₀₀ = 100(15) = 1500

Choice (2)

13. If m n and p are arithmetic progression then the m th term and pth term of arithmetic progression are in

Arithmetic progression
Geometric progression
Not neccessarly in arithmetic progression or geometric progression

Answer: Given m,n, and pare in A.P.

Let n-m =p- n = k

n = m + k,p = m + 2k

consider an A.P.

CA Foundation Maths Solutions For Chapter 6

a₁,a₂,a₃,a₄,…….whose connon difference is d.

a_m + d = a_m + 1[a_m is mth term]

a_m+ kd = a_m+k = a_n m+k

a_m + kd = a_m+2k = a_p m+2k

a_p – a_n = [a_m + 2kd]- (a_m + kd] = kd

a_n – a_m = a_p – a_n = kd ∴ a_m ,a_n and a_p are in A.P

m term, nth term and pth term are in A. P

Choice (1)

14. If m . n and p are in arithmetic progression then m th term, n th term and pth term of a geometric progression are in

Arithmetic progression
Geometric progression
Not neccessarly in arithmetic progression or geometric progression

Answer: Given m, n and p are in A.P.

Let n-m=p-n=k

n = m + k,p = m + 2k

Consider a G.P

g1,g2,g3——— whose common ratio is r.

gm(r) = gm+1 [gm is mth term]

gm(rk)gm+k = gn [m+k=n]

gm(r2) = gm+2k = gp [m+2k=p]

\(\frac{g_p}{g_m}=\frac{g_m r^{2 k}}{g_m r^k}=r^{-k}\) \(\frac{g_n}{g_m}=\frac{g_p}{g_{11}}=r^k\)

gm,gn and gp are in G. P.

Choice (2)

15. If sum of first 51 terms of an arithmetic progression is zero , then which of the following terms is zero?

13th
26th
17th
Cannot be determined

Answer: In an A.P. if the sum of the first n terms is Zero, and dn is odd, the

\(\frac{(n+1)^{t h}}{2}\) term is zero.

\(\left(\frac{51+1}{2}\right)^{t h}\)

26th term is zero

Choice (2)

16. If the sum of first 20 terms of an arithmetic progression is 30 and the sum of first 50 terms is also 30 then what is the sum of 21st term and the 50th term?

0
30
15
Cannot be determined

Answer: Given S₂₀=30 and S₅₀ = 30

S₅₀ = S₂₀ + t₂₁+——-+t₅₀

30 = 30 + t₂₁ +————-t₅₀

t₂₁+t₂₂——–+t₅₀=0

the mean of t₂₁ and t₅₀ is zero.

choice (1)

17. If the positive numbers m, n and p are in geometric progression then Ion gm, logn and log p are in

Arithmetic progression
Geometric progression
Cannot be determined

Answer: Given m, n and p are in G. P.

\(\text { Let } \frac{n}{m}=\frac{p}{n}=r\)

n = mr and p = mr²

Let log m=a…..(1)

Log n == log (mr)

= log m + log r[ log(xy) = logx + logy]

Let log r=b logn = a + b…..(2)

Log p = log (mr)²log m+ log r² =log m +2 log r [ log r² = 2logr]

∴ log p = a + 2b…..(3)

From 1,2,3

Log m, log n and log p are in A.P.

Choice (1)

18. If the geometric mean of two distinct positive numbers is 4, then the arithmetic mean of these two numbers is

<4
=4
>4
Cannot be determined

Answer: For any two unequal positive numbers a and b

\(\frac{a+b}{2}>\sqrt{a b}\)

Given √ab =4

Arithmetic mean, \(\frac{a+b}{2}>4\)

Choice (3)

19. What is the seventh term of a geometric progression whose first term is 3 and common ratio is 2?

Answer: In a G.P.

T_n = ar^n-1

T₇ = 3(2)^7-1

= 3(64) = 192

Choice (3)

20. What Is the product of first 9 terms of a geometric progression’ having a total of 13 terms given that 51th term is 2?

512
32
16
Cannot be determined

Answer: The 5th term is equidistant from the first term and the 9th term

5th term is the geometric mean of first and ninth term and moreover, it is the geometric mean of first 9 terms the product of the first n terms in a G.P = (Geometric mean of first n
terms)ⁿ

Product = 2⁹ = 512

Choice (1)

21. What is the geometric mean of the geometric progression having a total of 13 terms given the 7lh term is 4?

2
4
16
Cannot be determined

Answer: In G.P. if the number of terms is n which is odd, then the middle term i.e. \(\left(\frac{n+1}{2}\right)^{t h}\) term is the geometric mean.

\(\left(\frac{13+1}{2}\right)^{\text {th }}\) term i. e. the 7th term is the geometric mean i. e. the geomeric mean is 4.

Choice (2)

22. What is the sum of infinite geometric series 1,1/2,1/4,1/8,1/16,….?

\(1 \frac{255}{256}\)
2
3
4

Answer: Common ratio = \(\frac{1 / 2}{1}=\frac{1}{2}\)

\(S_a=\frac{a}{1-1}=\frac{1}{1-1 / 2}=2\)

Choice (2)

CA Foundation Maths Solutions For Chapter 6

23. Every number of an infinite geometric progression of positive terms is equal to m times the sum of the numbers is equal to m times the sum of the numbers that follow it. What is the common ratio of the progression?

\(\frac{m}{m+1}\)
\(\frac{1}{m+1}\)
\(\frac{2}{m+1}\)
Cannot be determined

Answer:

Let the geometric progression be a, ar, ar², …. given a = m \(\left(\mathrm{ar}+\mathrm{ar}^2+\cdots=\frac{\mathrm{ar}}{1-\mathrm{r}}\right.\)

=>mr = 1-r ⇒ mr + r = 1

\(\mathrm{r}=\frac{1}{m+1}\)

Choice (2)

24. What is the sum to 7 terms of a geometric progression whose first term is 1 and the 4th term is 27?

1093
2186
3279
4372

Answer: Given a= 1,t4= 27

t₄ = ar³

27 = (1)r³->r = 3

\(S_7=\frac{1\left(1-3^7\right)}{1-3}=\frac{1-2187}{2}=\frac{-2186}{-2}=1093\)

Choice (1)

25. What is the sum of the cubes of first (9 natural numbers?

45
2025
91125
Cannot be determined

Answer: Sum of the first 9 natural numbers = \(\frac{9(9+1)}{2}=45\)

Sum of the cubes of first n natural numbers = (sum of first n natural numbers)²

∴ Sum of the cubes of first 9 natural numbers =45²=2025

=2025

Choice (2)

26. In an arithmetic progression, the first and the last terms are 9 and 96 respectively if the sum of all the terms of the progression is 4200, what is the common difference ?

80
\(\frac{76}{79}\)
70
\(\frac{87}{79}\)
\(\frac{83}{79}\)

Answer:

Let the number of terms in the arithmetic progression be n.

\(\frac{n}{2}[9+96]=4200 \Rightarrow n=80 \text {. }\)

21th term = a+20d where a=9.

96= 9+(n-1)d = 9 + 79d

\(\mathrm{d}=\frac{87}{79}\)

Choice (4)

27. The sum of the first 30 terms of an arithmetic progression is 4500 the ratio of the first 20 terms and the last 10 terms in 4:5 find the first term.

Answer: Let the first term and the common difference of the arithmetic progression be a and d respectively.

\(\frac{30}{2}[2 a+29 d]=4500\) 2a + 29d = 300

Sum of the first 20 terms= \(\frac{4}{9}(4500)=2000\)

\(\frac{20}{2}[2 a+19 d] 2000\)

2a = 19d = 200

Subtraction the second equation from the first 10d = 100=>d = 10

First term = \(\frac{300-29 d}{2}=5\)

Choice (4)

28. The sum of all the three digit numbers which leave a remainder of 3 when divided by 7 is

64215
70720
64320
70821
64160

Answer: The first three-digit number when divided by 7 leaving a remainder of 3 is 101 and the last three-digit number when divided by 7 leaving a remainder of 3 is 997.

101 = 7 X 14 + 3

997 = 7 X 142 + 3

Number of 3 digit multiples of 7= \(\frac{997-101}{7}+1\) = 129 the sum of all these numbers =

\(\frac{129}{2}[101+997]=129(549)=70821\)

Choice (4)

29. The sum of all the natural numbers from 200 to 600 (both inclusive ) which are neither divisible by 8 nor by 12 is

1,23,968
1,33,068
1,33,268
1,87,332
1,34,168

Answer: Let the sum of all the natural numbers from 200 to 600 which are divisible neither by 8 nor by 12 be P.

P= sum ofall natural numbers from 200 to 600 – (sum ofall natural numbers divisible by either 8 or 12)

=sum of the first 600 natural numbers- sum of the first 200 natural numbers – (sum of all 3 digit natural numbers from 200 to 600 divisible by 8= sum of all 3 digit
natural numbers from 200 to 600 divisible by k12 – sum of all 3 digit numbers from 200 to 600 numbers divisible by both 8 and 12)

\(=\frac{(600)(601)}{2}-\frac{(199)(200)}{2}\)

– (200+208+…..+600)

+ (204+212+…. +600)

– (216+240+….+600) =(180300-19900)

\(-\left[\frac{51}{2}(200+600)+\frac{34}{2}(204+600)-\frac{17}{2}(216+600)\right]=133268\)

Choice (3)

30. The sum of four terms in arithmetic progression is 64. The sum of the squares of the first and the last term is 64 more than the sum of the squares of the second and the third terms. Find the first term, if the second term is less than third term.

Answer: Let the four numbers be a-3d, a-d, a + d and a-3d. a -3d+a-d+a+d+a+3d=64

4a = 64

a=16

(a- 3d)² + (a + 3d)² = (a- d)² + (a + d)² + 64

=> 18d² = 2d² + 64

d=√4 = 2 as d > 0

The first term = a- 3d = 10.

Choice (5)

CA Foundation Maths Sequence And Series Practice Problems

31. There are four distinct numbers in a sequence such that the first and the last terms are equal, the first three numbers progression and the last three are in geometric progression, find the common ratio of the last three numbers.

1
-1
-1/2
-2
2

Answer: Let the second, third and fourth numbers which are in geometric progression be a, ar and ar² respectively the first number = the fourth number = ar²

Since the first three are in arithmetic progression, ar², a and ar are in arithmetic progression.

=>2a =ar²+ ar

As the numbers are distinct, a≠0 and r≠1

=> 2=r²+r=>r²+r-2=0=>r=1 or -2 since r≠1,r=-2

Since r≠1,r=-2

Choice (4)

32. Four times the ninety first term of an arithmetic progression is equal to 5 times its eighty first terms. If the thirty first term of the progression is -630, find its fifty first term.

Answer: Let the first term and the common difference of the arithmetic progression be a and d.

4(a+90d) = 5 (a+80d)=>-40d=a

Thirty-first term of the at arithmetic progression =a +30d= -630 d=+63.

Fifty first term of that arithmetic progression =a+50d=-40d+50d=+10d=630

Choice (3)

33. The ratio of the sum of the first n terms of two arithmetic progressions is given by 5n+4:8n-15. Find the ratio of the 12th terms of the two arithmetic progressions.

125:161
119:169
123:166
115:169
119:238

Answer: Let the two arithmetic progressions’ have first terms and common differences as a₁,d₁ and a₂,d₂ respectively. Ratio of the sum of their first n terms

\(=\frac{\frac{n}{2}\left[2 a_1+(n-1) d_1\right]}{\frac{n}{2}\left[2 a_2+(n-1) d_2\right]} \quad \frac{a_1+\frac{n-1}{2} d_1}{a_2+\frac{n-1}{2} d_2}=\frac{5 n+4}{8 n-15} \ldots \ldots\)

Ratio of their 12th terms = \(\frac{a_1+11 d_1}{a_2+11 d_2}\)

Substituting \(\frac{n-1}{2}=11\) in the L.H.S. we get ratio of their 12th terms.

\(=\frac{5(23)+4}{8(23)-15}=\frac{119}{169}\)

Choice (2)

CA Foundation Maths Solutions For Chapter 6

34. The number of terms of the series 40,37, 34,… for which the sum is 282 is _________

Answer: The first term of the arithmetic progression series is 40 and the common difference is -3 let the number of terms in the series for which the sum is 282 be n

\(\frac{n}{2}[2(40)+(n-1)(-3)]=282\) \(\frac{n}{2}[83-3 n]=282\)

3n²- 83n + 564 = 0

3n(n-12)-47(n-12)=0

(n-12)(3n-47)=0

n=12 or 47/3 as n is an integer, n = 12.

Choice (5)

35. The sum of the first, sixth, eighth, ninth and thirteenth terms of an arithmetic progression is 243. Find the sum of the first 13 terms of the arithmetic progression.

1134
1296
1458
526.5
1588

Answer: Let the first term and the common difference of the arithmetic progression be a and d respectively.

a + (a + 4d) + (a + 5d) + (a + 7d) + (a + 8d) + a + 12d) = 243

6a = 36d = 243……. (1)

Sum of the first 13 terms = [2a + 12d] = 13[a = 6d] \(=\frac{(13)(243)}{6}(\text { from }(1))=526.5[latex]

Choice (4)

36. If 5x+y,3x+2y and 3x+y are in arithmetic progression and 5x+1, 3(x+1) and 3x are in geometric progression, find (x,y).

(4,3)
(5,2)
(3/2,3/2)
(3,3)
(3/4,3/4)

Answer: 2(3x + 2y)=5x + y + 3x + y

6x + 4y = 8x + 2y=>x = y

[3(x + 1)]² = (5x + 1)3x

9x² + 18x + 9 = 15×2 + 3x

6x²- 15x- 9 = 0

Dividing both sides by 3

2x²- 5x- 3 = 0

(2x =1)(x-3) = 0

X=-1/2 or 3; going by the choices, x=3,

Choice (4)

37. The maximum value of the sum of the series series 50,46,42,…is

Answer: Let the number of terms of the series which give the maximum sum be n

Sum =n/2[2(50)+(n-1)(-4)]

=n/2[104-4n]=n[52-2n]=2n[26-n]

When the sum of two quantities is constant, their product is maximum when the quantities are equal as n and 26-n have a constant sum of 26 their product is maximum when n = 26-n

26/2=13=n

Maximum sum = (2)(13)(13)=338

Choice (5)

38. In a geometric progression the first three terms are 10n +36,6ct + 12 and 4a respectively, find the sixth term of the geometric progression if

a > 0
5
2
3
9
6

Answer: (6n + 12)² =(10a+36) 4a

36a² + 144a + 144 = 40a² + 144a

144 = 4a², a = 6 as a > 0

sixth term of the G. P = [latex](10 a+36)\left(\frac{6 a+12}{10 a+36}\right)^5=96\left(\frac{48}{96}\right)^5=3\)

Choice (3)

39. A ball dropped from 36 m above the ground rebounds to 1/3rd of the height it falls from. If it continues to rebound in this manner, find the total distance the ball can cover.

96m
72m
54m
76m
108m

Answer:

the height to which the ball rebounds after the first fall = 36 (1/3)=12m the 2nd time it falls from 12m height. The distance the ball rebounds after the second time it falls =

36(1/3)(1/3)= 4m, it falls from 4 m height

Likewise the distance rebounded by the ball after every time can be found to be (1/3) of distance rebounded after the previous time it fell

Total distance it covers, 36+12+12+4+4+…a= (36+12+4+…a)+(12+4+…a)

=(36+12+4+…..a)+(12+4+…..a)= \(\left(\frac{36}{1-1 / 3}\right)+\left(\frac{12}{1-1 / 3}\right)\)

54 + 18=72m.

Choice (2)

40. Three distinct numbers in geometric progression have a product of 1728 the sum of the products taking two numbers at a time is 456. Find the least number.

Answer: Let the numbers in geometric progression be a/r,a and ar(a/r)(a)(ar)=1728

\(a=\sqrt[3]{1728}=12\) \(\frac{a}{r}(a r)+a(a r)+\frac{a}{r}(a)=456\) \(a^2\left(1+r+\frac{1}{r}\right)=456\) \(\frac{r+r^2+1}{r}=\frac{456}{12^2}=\frac{19}{6}\)

6r+6r² + 6 = 19r

6r² -13r + 6 = 0

3r(2r- 3)- 2(2r- 3) = 0

(3r-2) (2r-3)=0, r= 2/3 or 3/2

The least number = \(\frac{12}{3 / 2}\)

Choice (1)

41. The value of sum of first 75 terms of the sequence 150 x 2 + 148 x 4 + 6146 x 6 + is

252850
292600
273950
284050
231300

Answer: Given series is 150 x 2 + 148 x 4 + 146 x 6 +….75 terms

The numbers 150, 148, 146 … from an arithmetic progression, with a=150, d=-2.

The numbers 2, 4, 6…… from an arithmetic progression with a=2, d=2

Hence, tn then nth term of the series is:

[150 + (n- 1)(—2)][2 + (n- 1)2] = (152- 2n)2n.

t_n = 304n – 4n²

Hence sum to 75 terms is s75 = 304(1 + 2 + 3 …+ 75)- 4(1² + 2² + 3² + 75²)

\(=304 \times \frac{75 \times 76}{2}-4 \times \frac{75 \times 76 \times 151}{6}=292600\)

Choice (2)

42. The sum of the first 20 terms of the series \(\frac{3}{2}+\frac{15}{4}+\frac{47}{8}+\cdots \text { is }\)

\(\frac{\left(2^{20}\right)(420)-2^{20}+1}{2^{20}}\)
\(\frac{\left(2^{20}\right)(210)-2^{20}+1}{2^{20}}\)
\(\frac{\left(2^{20}\right)(210)+2^{20}+1}{2^{20}}\)
\(\frac{\left(2^{20}\right)(420)+2^{20}+1}{2^{20}}\)
\(\frac{\left(2^{20}\right)(420)-2^{20}-1}{2^{20}}\)

Answer:

\(\frac{3}{2}+\frac{15}{4}+\frac{47}{8}+\cdots \text { for } 20 \text { terms }\) \(=2-\frac{1}{2}+4-\frac{1}{4}+6-\frac{1}{8}+\cdots \text { for } 20 \text { terms }\)

=2+4+6+…for 20 terms— \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\right)\) for 20 terms = 2(1 + 2 + 3+…. +20)

\(\frac{\frac{1}{2}\left(1-\left(\frac{1}{2}\right)^{20}\right)}{1-\frac{1}{2}}=\left[\frac{(420)(2)^{20}-2^{20}+1}{2^{20}}\right]\)

Choice (1)

43. The maximum number of terms common to the arithmetic progression 3,7, 11, 15, 19,23,…403 and 5,11,17,23,29,35,… 505 is

Answer: The first common term to both arithmetic progressions is 11 and the common differences of the first and second arithmetic progressions are 4 and 6 respectively the next term common to both arithmetic progressions is 23 = 11+ L.C.M, of common differences of the two progressions= 11+12.

It can be confirmed that the term common to both series have a common difference of 12 the last term common to both progressions is less than minimum of last terms of both
progressions i.e. 403. Let n terms be common to both progressions.

The last term common to both arithmetic progressions is 11 + 12(n – 1). This is less than 403.

\(n-1<\frac{(403-11)}{12}=32 \frac{2}{3}\)

\(n<33 \frac{2}{3}\) Hence n must be a maximum of 33.

44. Find the value of \(\frac{1}{3^2-2^2}+\frac{1}{7^2-2^2}+\frac{1}{11^2-2^2}+\) \(\cdots+\frac{1}{39^2-2^2}\)

5/41
31/41
18/37
19/37
10/41

Answer:

\(\frac{1}{3^2-2^2}=\frac{1}{(3-2(3+2)}=\frac{1}{4}\left[1-\frac{1}{5}\right]\) \(\frac{1}{7^2-2^2}=\frac{1}{(7-2)(7+2)}=\frac{1}{4}\left[\frac{1}{5}-\frac{1}{9}\right]\) \(\frac{1}{39^2-2^2}=\frac{1}{(39-2)(39+2)}=\frac{1}{4}\left[\frac{1}{37}-\frac{1}{41}\right]\)

the required sum is 1/4[1-1/41}=10/41

Choice (5)

CA Foundation Maths Solutions For Chapter 6

45. If an is defined as follows, find the value of a50

1. a_n = 1,if n = 0

2. a_n = a_n – 1,if n = 3k, k being a positive integer

3. a_n = 3 a_n-1 If n = 3k + 1, where k is a whole number

4. a_n = 2 a_n-1 if n = 3k + 2, where k is a whole number

6¹⁷
3¹²2¹⁷
3¹⁷2¹⁸
6¹⁸
none of these

Answer: Tabulating the values of a, we have a₀ = 1

a₁ = 3

a₂ = 2(3)

a₃ = 2(3)

a₄ = 2(3)²

a₅ = 22(3)²

a₆ = 22(3)²

a₇ = 22(3)³

a₈ = 23(3)³

46. the first 15 terms of the series 32.1 + 42. 2 + 52. 3 + 62. 4 + 72. 5 +…..is _____________

16120
24420
21840
20840
19840

Answer: Each of the terms being added is of the form (a + 2)² a where 1 <a <15

Their sum =\( \sum_{a=1}^{15}(a+2)^2 a=\sum_{n=1}^{15} a^3+4 a^2+4 a\)

\(=\left(\frac{a(a+1)}{2}\right)^2+4\left(\frac{1}{6}\right) a(a+1)(2 a+1)+\frac{4 a(a+1)}{2}\) \(=\left(\frac{15(16)}{2}\right)^2+4\left(\frac{1}{6}\right)(15)(16)(31)+\frac{4(15)(16)}{2}\)

=(120)² + (4)(40)(31) + 480 = 19840

Choice (5)

47. 3100 – 2(396 + 397 + 398 + 399) =

3⁹⁶
2(3⁹⁶)
3⁹⁷
4(3⁹⁶)
5(3⁹⁶)

Answer:

3¹⁰⁰ – 2(3⁹⁶+ 3⁹⁷ + 3⁹⁸ + 3⁹⁹)

= 3¹⁰⁰– (2)3⁹⁶(3⁴ – 1)/(2) = 3⁹⁶

Choice (1)

48. A Eugenics research worker is working with two types of bacteria. One type triples in number every ten minutes while the other one becomes 5 times in the same time, if the total number of bacteria after half an hour is 1118, what was the total numberat the beginning?

Answer: Let there be x bacteria of the first kind and y of the second.

27x + 125y = 1118

The remainder of 1118/27 is 11

Taking successive multiplies of 125, the remainders are 17,7,24,14,4,21,11 (Add 17 or subtract 10)

27(9)+125(7)=1118

Hence 27(9+125)+125(7-27) and other such expression are also equal to 1118, but x=9, y=8 is the only solution in positive integers.

At the beginning there were 9+7=16 bacteria.

Choice (3)

49. The sum to infinity of the series a² (a² + b²) + a⁴(a⁴ + b⁴) + a⁶(a⁶ + b⁶) +…. where |a|≤1 and |b| ≤1 is

\(\frac{a^2 b^2+a^4+2 a^6 b^2}{\left(1-a^2 b^2\right)\left(1-a^4\right)}\)
\(\frac{a^2 b^2-a^4-2 a^6 b^2}{\left(1-a^2 b^2\right)\left(1-a^4\right)}\)
\(\frac{a^2 b^2-a^4+2 a^6 b^2}{\left(1-a^2 b^2\right)\left(1-a^4\right)}\)
\(\frac{a^2 b^2+a^4+2 a^6 b^2}{\left(1-a^2 b^2\right)\left(1-a^4\right)}\)
\(\frac{a^2 b^2-a^4-2 a^6 b^2}{\left(1-a^2 b^2\right)\left(1-a^4\right)}\)

Answer:

The series when expended becomes

a⁴ + a²b² + a⁸ + a⁴b⁴ + a¹² + a⁶b⁶ + …..∞

= a⁴ + a⁸ + a¹² +……∞

\(=\frac{a^4}{1-a^4}+\frac{a^2 b^2}{1-a^2 b^2}=\frac{a^4\left(1-a^2 b^2\right)+a^2 b^2\left(1-a^4\right)}{\left(1-a^4\right)\left(1-a^2 b^2\right)}=\frac{a^4-2 a^6 b^2+a^2 b^2}{\left(1-a^4\right)\left(1-a^2 b^2\right)}\)

Choice (1)

50. If S=3+7x+llx2 + 15×3 + …..(3 +4n)xn…and |x| < 1,S =

\(\frac{3+x}{(1-x)^2}\)
\(\frac{3+x}{1+x^2}\)
\(\frac{3-x}{(1-x)^2}\)
\(\frac{3-x}{(1+x)^2}\)
none of these

Answer:

S=3+7X+11x² + 15x³ +……-> (1)

Sx = 3x + 7x² +11x³+…..-> (2)

(1)- (2)

5(1- x) = 3 + 4x + 4x² + 4x³ + ……..

= -1+4/1-x = x + 3/1-x

S = (x + 3)/(1- x)²

Choice (1)

51. Chidanand and Dharmesh joined a company on 1 jan 1991. Chidanand was offered a starting salary of Rs. 3000 p.m. with an kannual increment of Rs.500 p.m. Dhamesh was put on a starting salary of RS.2000 p.m. with a six monthly increment of Rs. 350 p.m. if they continued working till 31 December 2000, how much more or less did Dharmesh receive compared to chidanand?

Rs.9000 more
Rs.9000 less
Rs.7500 more
Rs.7500 less
Rs. 4500 more

Answer: Chidanand’s first monthly salary is Rs.3000

His last monthly salary is Rs.7500

His average monthly salary is Rs.5250

Dharmesh’s first monthly salary is Rs.2000

His last monthly salary is is Rs 8650

His average monthly salary is Rs 5325

Dharkmesh gets 10(12) (5325-5250) = Rs. 9000 more than chidanand.

choice (1)

Chapter 6 CA Foundation Maths Answer Key

52. The sum of the first n terms of an arithmetic progression is p. the sum of the first 2n terms is Q. if Q= (4/P, find the sum of the first 3n terms of the arithmetic progression

Answer:

The sum of the first n terms is p₁ (say),that of the next n terms is P₂(say) and that of the third set of n terms is P₃ (say)

P₁, P₂, P₃ Have to be in arithmetic progression.

As P₃ are in arithmetic progression.

P₃are in aritmetic progression. P₃ = 2(Q- P)- P = 2Q- 3P …. (1)

The sum of the first 3n terms is P₁ + P₂ + p₃ = Q+P₃ = Q + (2Q- 3P)from (1) = 3Q- 3P.

As Q =4/3p,P₁ + P₂ + P₃ = 4P-3P = P

Choice (4)

53. The ratio of the sum of first 20 terms of an arithmetic progression and the sum of its first 30 terms is 10:31 if the common difference of the arithmetic progression is 2, find the first term.

\(-6 \frac{3}{8}\)
\(-9 \frac{7}{8}\)
\(-8 \frac{3}{8}\)
\(-10 \frac{5}{8}\)
\(-9 \frac{5}{8}\)

Answer: Let the first term and the common difference of the terms in arithmetic progression be a and d respectively

\(=\frac{\frac{20}{2}[2 a+19 d]}{\frac{30}{2}[2 a+29 d]}=\frac{10}{31} \Rightarrow \frac{2(2 a+19 d}{3(2 a+29 d)}=\frac{10}{31}\)

124a + 117d = 60a + 870d

64a = -308d

a=-308/64(2)

\(-9 \frac{5}{8}\)

Choice (5)

54. Find the sum (1)(3)+ (3)(5)+ (5)(7) + (7)(9) + …………………………………. up to 50 terms.

171700
171850
1717750
1701650
171650

Answer: (1)(3)+(3)(5)+(5)(7)+(7)(9)+……

The first parts of each term 1.3,5,7,9,… are in arithmetic progression and its nth term is (2n-1) the second parts of each term 3,5,7,9,1 1,… are also in arithmetic progression and its nth term is (2n+1)

the n term of the series is (2n-1) (2n+1) =4n² – 1

the sum ofn term of the series is (4n2-1)

= Σ(4n²- 1) =4 ∑ n² – ∑1 =\(\frac{4 n(n+1)(2 n+1)}{6}-n\)

The sum of the first 50 terms of the series \(=\frac{4 \times 50 \times 51 \times 101}{6}-50\)

=100 x 17 x 101- 50

= 171700-50=171650

Choice (5)

55. Find the sum \(\frac{1}{2}+\frac{2^2}{2^2}+\frac{3^2}{2^3}+\frac{4^2}{2^4}+\)

\(\frac{5^2}{2^5}+\frac{6^2}{2^6}+\cdots\)

Answer:

\(\text { let } S=\frac{1^2}{2}+\frac{2^2}{2^2}+\frac{3^2}{2^3}+\frac{4^2}{2^4}+\frac{5^2}{2^5}+\frac{6^2}{2^6}+\cdots\) \(\frac{S}{2}=\frac{1}{2^2}+\frac{2^2}{2^3}+\frac{3^2}{2^4}+\frac{4^2}{2^5}+\frac{5^2}{2^6}+\cdots\) \(S-\frac{S}{2}=\frac{S}{2}=\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{1}{2^4}+\frac{9}{2^5}+\frac{11}{2^6}+\cdots\) \(\frac{S}{4}=\frac{1}{2^2}+\frac{3}{2^3}+\frac{5}{2^4}+\frac{7}{2^5}+\frac{9}{2^6}+\cdots\) \(=\frac{1}{2}+\left[\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots\right]\) \(=\frac{1}{2}+\frac{1 / 2}{1-1 / 2}=\frac{1}{2}+1=\frac{3}{2} \Rightarrow S=\frac{3}{2} \times 4=6\)

56. A total of 810 balls are arranged in N layers. The ith layer from the top where 2 < i < N has 4 more balls than the layer immediately above it. Which of the following is not a possible value of N?

Answer: The number of balls in the 1st, 2nd … nth layer from the top be f let the top are in arithmetic progression. Let the number of balls in the 1st layer from the top be f
total number of balls = N/2(2f+ (N – 1)4)

810 = N(f+ (n- 1)2)

To satisfy the above equation, N must be a factor of 810 only choice (5) violates this condition

Choice (5)

57. X is the set of the first 1990 natural numbers. A is an arithmetic progression having at least 3 elements. First element is the least element of x and its last element is the greatest element of x How many possibilities does a have?

Answer: The least element of A is 1 and its greatest element is 1990 suppose A has N elements and a common difference of D. then 1990= 1+(N-1) D (N-1)D=1989 =3²(17) (13)——(1)
N ≥ 3

1989 has (3) (2) (2) or 12 factors or 6 pairs of factors.

(N-1)D=1(1989)= 3(663)=9(221)=13(153)=17(117)=39(51)

As N ≥ 3, ,N-1 ≥ 2

N-1 can be 3,9,13,17,39,51,117,153,221,663 or 1989 i.e. N can have 11 values.

Choice (2)

58. Find the number of common terms to the sequences 7,11.15,19,23…..411 and 8,13,18,23,…..463

Answer: The two equerries are Aps. For any two Aps, their common terms also from on AP.

Moreover, the common difference of that AP is the LCM of the common differences of the Aps the common difference of the first and second Aps are 4 and 5 respectively the AP of common terms has a common difference of 20.

First common term to the two Aps = 23 suppose the AP of common terms has n terms then the nth cannot exceed min (last terms of the Aps) i.e. min (411,463)=411. Also N must be the largest value satisfying this condition.

∴ 23 + 20(N- 1) ≤ 411

20N = 3 ≤ 411

N ≤ 20.4

N = 20

Choice (5)

59. The nth term of the series 6,13,20,Is 636. Find n.

Answer: The nth term of an arithmetic progression with the first term a and common difference d is given by a +(n-1) d. for the given series.

a=6 and d=7 its nth term =>6+(n-1)7=636⇒n=91

Choice (2)

60. The sum of sixty numbers in arithmetic progression is 7,800. The largest of the numbers is 12 times the smallest. Find the smallest number.

Answer: Let the smallest numberbe a and common difference of the arithmetic progression be largest number = 12a

Sum of the 60 numbers =60/2[a+12a]

390a=7800

a=20

Choice (5)

61. The sum of three numbers in arithmetic progression is 36.the sum of the squares of the three numbers is 464. Find the smallest number.

Answer: Let the three numbers be a-d a and a+d

a-d + a + a + d = 36.

3a=36 a=12

(12- d)² + 12² + (12 + d)² = 464

d² = 16

d=√16=±4

If d is either +4 or -4 the numbers are 8, 12,16 the smallest is 8.

Choice (1)

62. Find the sum of all the numbers divisible by 8 and lying between 300 and 950.

90,960
1,00,480
96,480
50,544
52,648

Answer: The first number divisible by 8 above 300 is 304= (8x 38) the last number before 950 divisible by 8 is 944 i.e.8 x 118

Total number ofmultiples between 304 and 944

\(=\frac{9 \cdot 4-30 \cdot}{8}+1=81, n=81 .\)

Sum of all these numbers 81/2(304- 944) = 50544.

Choice (4)

63. The seventh term of a geometric progression is 3645. If the product of its first 3 terms is 3375, find the first term of the geometric progression.

Answer: Let the first, second and third terms of the geometric progression be a/r, a and ar respectively

a/r a ar = a³ =3375 =>a=√3375 = 15

Seventh term of the geometric progression =\(\frac{a}{r} r^6\)= ar⁵(since first term is a/r) 15r⁵=3645

r⁵ = 243

r=⁵√243 = 3

a/r=5

Choice (1)

64. A geometric progression has a sum to infinity. The value of the cube of any term is 1 /8th of the cube of the sum of the terms which follow it . find the sixth term of the geometric progression if the first term is 4.

\(\frac{2}{3}\)
\(\frac{64}{81}\)
\(\frac{128}{243}\)
\(\frac{256}{729}\)
\(\frac{512}{2187}\)

Answer: Let the series be 4, 4r, 4r², 4r³

Given that,(4)³ = \(\frac{1}{8}\left\{\frac{4r}{1-r}\right\}^3\)

\(\Rightarrow 4=\frac{1}{2}\left(\frac{4 r}{1-r}\right)\)

r=2/3

the sixth term = ar⁵ = (4)(2/3)⁵=128/243

Choice (3)

65. There are three numbers in arithmetic progression having a sum of 51. If the first is increased by 2, the second is increased by 1 and the third is increased by 3, the numbers would be in geometric progression. Find the smallest number in the geometric progression.

Answer: Let the first second and third numbers in arithmetic progression be a- d a and a + d.

a-d + a = a = d = 3a = 51=>a = 17

a-d + 2,a + 1 and a + d + 3 i.e. (19- d) (20 + d) = 324

=> d² + d- 56 = 0=>d = -8 or 7

So, the numbers in geometric progression are either 27, 18 and 12 or 12, 18 and 27, so whatever be the case the smallest is 12.

Choice (2)

66. The product of three numbers in geometric progression is 1728. If their sum is 37, find the smallest number.

Answer:

Let (a/r) a and ar be the numbers in geometric progression. Given that (a/r) (a)(ar) = 37

=>12/r+ 12 + 12r=37=>12r² – 25r +12 = 0

=>(3r- 4)(4r- 3) = 0=>r = 3/4 or 4/3

the numbers are 9,12,16 or 16, 12, 9, in either case, the smallest is 9.

Choice (1)

67. A geometric progression has the sum to infinity 4. The sum to infinity of the sum of the squares of the terms of the geometric progression is 6. Find the first term of the geometric progression.

24/11
32/11
96/11
40/11
16/11

Answer:

Let the first term be a and the common ratio be r.

\(\frac{a}{1-r}=4 \ldots .\) \(\frac{a^2}{1-r^2}=6 \ldots \ldots(2)\) \(\Rightarrow \frac{a}{1-r}\left(\frac{a}{1+r}\right)=6\)

from (1)and (3), We have,

\(a=\frac{6}{4}(a+r)=(1-r) \Rightarrow \frac{11}{2} r=\frac{5}{2}\)

r=5/11 first term = 4(1-r)=24/11

Choice (1)

68. An infinite geometric progression having a finite sum has its first term as 12. The difference of the third and the fifth terms of the geometric progression is 9/4 if the common ratio is a rational number; find the sum to infinity of the terms of the geometric progression.

12 or 15
16 or 10
20 or 12
24 or 8
12 or 8

Answer:

Let the common ratio of the geometric progression be r. the third term of the geometric progression =12r²

The fifth term of the geometric progression 12r⁴

As the geometric progression has a sum to infinity, -1 < r < 1

Hence r² < r⁴ = 9/4

Let r = a

12(a- a²) =9/4

48a- 48a² -9 = 0

16a² – 16a + 3 = 0, dividing both sides by 3

16a² – 12a -4a + 3

4a (4a- 3)- 1(4a- 3) = 0

(4a- 3)(4a- 1) = 0

a = 3/4 or 1/4

r=±1/2 as r is rational sum to infnity= \(\frac{12}{1 \pm \frac{1}{2}}\)

= 24 or 8

Choice (4)

69. Find the sum of the squares of the first 12 terms of the arithmetic progression for which the sum of the first n terms is 3n² + 6n

12450
13170
14100
26316
24216

Answer:

Given s_n = n(3n + 6)

a = T₁ = S₁ = 1(3 + 6) = 9

T₂ = S₂– S₁ = 2(6 + 6)- 9 = 15

d=T₂– T₁ = 15- 9 = 6

the series is 9,15,21,….

Sum of the squares of the first 12 terms of the arithmetic progression

=9² + 15² + 21²+ …….+ 75²

=3²(3²+5²+7²+…..25²)

=3²(1²+ 2²+ 3² +4² +…..+ 25²)+- 3²(1²)

=3²[(1² + 2² + 3² + 4² +….. 26²)- (2² + 4² +……+ 26²)- 1²]

\(=9\left[\frac{(26)(27)(53)}{8}-\frac{2^2(13)(14)(27)}{6}-1\right]\)

=9[117(53)-4(63)(13)-1]

=9(6201-3276-1]

=9[2924]=26316

Choice (4)

70. The sum of the first 12 terms of log31/2X + log31/4 x log31/6X + … 312 find x.

Answer:

\(\log \frac{1}{3^2} x+\log \frac{1}{3^4} x+\log \frac{1}{3^6} x+\cdots\) \(\frac{\log _3 x}{\log _3 3^{1 / 2}}+\frac{\log _3 x}{\log _3 3^{1 / 4}}+\frac{\log _3 x}{\log _3 3^{1 / 6}}+\cdots \text { for } 12 \text { terms }\)

(Rewriting each term being added which is in the form

\(\log _{\frac{1}{b^2}} a a s \frac{\log a}{\log b^{\frac{1}{2}}}\)

and taking the base for both numerator and denominator as 3)

\(\frac{\log _3 x}{\frac{1}{2} \log _3{ }^3}+\frac{\log _3 x}{\frac{1}{4} \log _3{ }^3}+\frac{\log _3 x}{\frac{1}{6} \log _3 3}+\cdots \text { for } 12 \text { term }\)

= 2log₃ x +4log₃ x +6log₃ x +8log₃ x + + 24log₃ x

=(2+4+6+….+24) log₃ x= 156(log₃ x) = 132

=> log₃ x= 2=> x= 3² = 9

Choice (1)

71. If p² = q² -q⁴ + q⁶ – q⁸ +….. ∞, find P given |q| < 1

\(\frac{ \pm q}{\sqrt{1-q^2}}\)
\(\frac{ \pm(q+1)}{\sqrt{q^2+1}}\)
\(\frac{ \pm q}{\sqrt{q^2-1}}\)
\(\frac{ \pm q}{\sqrt{q^2+1}}\)
\(\frac{ \pm(q-1)}{\sqrt{q^2+1}}\)

Answer:

Given p² = q²- q⁴ + q⁶– q⁸ +….is an infinite geometric progression with |q| <1 =>|q²| < 1. the common ratio of the geometric progression is -q²

\(p^2=\frac{q^2}{1-\left(q^2\right)}=\frac{q^2}{1+q^2} \Rightarrow p=\frac{ \pm q}{\sqrt{1+q^2}}\)

Choice (4)

72. There are three numbers in geometric progression. When the middle number is tripled. The numbers will be in arithmetic progression. If the common ratio is greater than 1, find its value.

4+3√2
3+4√2
3√2-4
3+2√2
6+3√2

Answer: Let the three numbers be a, 3ar and ar² therefore a, 3ar and ar2 are in arithmetic progression

2(3ar) = a + ar²

r² – 6r + 1 = 0

\(r=\frac{6 \pm \sqrt{32}}{2}\)

3±2√2

as r > 1,r = 3 + 2√2 Choice (4)

CA Foundation Sequence And Series Solved Examples

73. The 10th term of an arithmetic progression is 15. II the sum of the squares of the 7th, 10th and the 13th terms is minimum, the common difference is_.

Answer: Let the first term and common difference of the arithmetic progression be a and d respectively given that a+9d=15.

(a+6d)²+(a+9d)²+(a+12d)²

(a+9d-3d)²+(a+9d)²+(a+9d+3d)²

(15-3d)²+15²+(15+3d)²

=3(15)²+18d² this is minimum, when d=0.

Choice (1)

74. The sum to infinity of a geometric progression is 39/2. The sum to infinity of squares of the terms of the geometric progression is 253.5. Find the sum to infinity of cubes of the terms of the geometric progression.

252.5
240.5
253.5
260.5
None of these

Answer:

Let the first term and common ratio of the geometric progression be a and r respectively

\(\frac{a}{1-r}=\frac{30}{2} \ldots \ldots\) \(\frac{a^2}{1-r^2}=253.5=\frac{507}{2} \ldots . .2\) \(\left(\frac{a}{1-r}\right)^2=\left(\frac{39}{2}\right)^2\)

Squaring (1 ) on both sides… (3)

Dividing (2)by (3)and simplifying

\(\frac{\frac{a^2}{(1-r)(1+r)}}{\frac{a^2}{(1-r)^2}}=\frac{\frac{507}{2}}{\frac{1521}{4}} \Rightarrow \frac{1-r}{1+r}=\frac{2}{3} \Rightarrow r=\frac{1}{5}\) \(\text { from }(1), a=\frac{39}{2}(1-r)=\frac{78}{5}\)

sum of the cubes of the terms of the GP

\(=\frac{a^3}{(1-r)^3}=\frac{\left(\frac{78}{5}\right)^3}{\left(\frac{1}{5}\right)^3}=\frac{(78)(78)(78)}{124}=\frac{78(39)(39)}{31} \text { which is }>1000\)

Choice (5)

75. Find the 50th term of the following series. 1 + 3 + 7 + 13 + 21 +….

2451
2561
2781
2341
2671

Answer:

S₅₀ = 1 + 3 + 7 + 13 +…… T₄₉ + T₅₀ … ( 1)

Also s₅₀ = 1 + 3 + 7 + 13 + …..T₄₉ + T₅₀ … (2)

(1)- (2) =>0 = (1 + 2 + 4 + 6 +…..50 terms)- T50

=> T₅₀ = 1 + (2 + 4 + 6 + ……49 term)

=> T₅₀ = 1(49)(50)/2=1 + 2450 = 2451

Choice (1)

76.The terms a₁, a₂……..a₁₇ in arithmetic progression. If a₃ +a₈ + a₁₁ + a₁₄ = 100, then a₁ + a₂+ +a₃……… a₁₇ =__________

Answer:

if1st term T₁ = a, then nth Term T_n = a + (n- 1)d.

Given a₃ + a₈ + a₁₁ + a₁₄= 100

a₁+ 2d + a₁ + 7d + a₁10d + a₁+ 13d = 100

=>4a₁+ 32d = 100=>2a₁+ 16d = 50……(1)

To find: a₁+ a₂ + a₃ +…… a₁₇

= n/2 [first term + last term | =>17/2 |a₁ + a₁₇| = 17/2 |a₁ + a₁₇ + 16d]

=17/2[2a₁ + 16d] = 17(50)/2 = 425

Choice (4)

77. If five distinct numbers a, b, c, d, e are in arithmetic progression with common difference of ‘D’, then a- 3b+5c-5d+2e is equal to D

Answer:

Let x be the common difference

=>b = a + x, c = a + 2x, d = a + 3x and e = a + 4x.

a- 3b + 5c- 5d + 2e

= a- (3a + 3x) + (5a + 10x) – (5a + 15x) + 2a + 8x)

= (a — 3a + 5a- 5a + 2a) + (-3x + 10x- 1 5x + 8x) =0 + 0 = 0

Choice (4)

78. If in a geometric progression the fifth term is 8 times the second term and the sum of the first, third and sixth terms Is 111, what is the 7th term?

Answer: Let the first term of GP= A and common ratio = R

=>AR^N-1 = Nth term =>AR⁴= 8(AR)

=>R³ = 8=>R = 2 ……..(1)

A+AR² + AR⁵ = 111

=>A{1 + R² + R⁵) = 111…..(2)

Put R =2 in (2) => A( 1 + 4 + 32) = 11 1 => A =111/37 = 3

7th Term = AR⁶=3x (2⁶)=192

Choice (5)

79. If the sum of \(1+\frac{2}{x}+\frac{4}{x^2}+\frac{8}{x^3}+\cdots \infty\) is finite and x>0, then which of the following is true?

X> 2
X < 2
X > 1/2, x < 2
X < 1/2
1/2 < x < 2

Answer:

Since the sum \(1+\frac{2}{x}+\frac{4}{x^2}+\cdots \infty\) is finite, its common ratio should be less than 1.

\(S_{\infty}=\frac{a}{1-r}\)

common ratio = \(\frac{2}{\frac{x}{1}}=\frac{2}{x}\)

But r < 1

\(\Rightarrow \frac{2}{x}<1 \Rightarrow \frac{x}{2}>1 \Rightarrow x>2\)

Choice (1)

80. In an arithmetic progression, the 10th term is 11 and the 11th term is 10, How many consecutive terms (starting from the first term) o f the arithmetic progression should be considered so as to make their sum equal to zero?

Answer:

T_n=a+(n-1)d

(a= first term,d= common difference, Tn = n th term)

T₁₀ = a+9d=11……………(1)

T₁₁= a+10d=10…………….(2)

solving (1) and (2)

d= – 1, a = 20

Say ‘n’ terms are needed so as to make sum =0

=>n/2[2a+(n-1)d]=0

=>n/2[40+(n-1)(-1)]=0

=>40 – n + 1 = 0=>n =41

first 41 terms of the arithmetic progression when added give the sum as zero

Choice (2)

81. The first, second and third terms of a geometric progression with distinct terms, are equal to the first, ninth and thirty-first terms, respectively of an arithmetic progression. If the first term of either progression is 11. Find the common ratio of the geometric progression.

Answer: Let the GP be 11,11r, 11r²,r=1

The 1st 9th and 31st terms or the arithmetic progression are 11, 11+8d,11+30d

11r = 11 + 8d=>11 (r- 1) = 8d …. (1)

11r² =11 + 30d =>11(r²-1 )= 30d….(2)

(2) + (1)=>r+= 1 15/4 =>r =11/4

choice (2)

82. The first term of an arithmetic progression consisting of integers, is the common ratio of a geometric progression. The first term of the geometric progression is the common difference of the arithmetic progression. The sum of the first 11 terms of the arithmetic progression is 341 and the sum of the first three terms of the geometric progression is 18. Find the sum of the common difference and the common ratio.

Answer: If r is the common ratio of the geometric progression and d is the common difference of arithmetic progression first term of arithmetic progression = r and that of geometric
progression =d.

11/2[2r +(11-1)d] = 341

=>r+ 5d = 31…. (1)

=>d+ dr +dr² = 18

As arithmetic progression consists of only integers,both r and d are integers, so, Geometric progression also consists of only integers, By trial and error we get r=1and d=6
r + d = 7

Choice (5)

83. The three terms a + 2b, b+2c are c +2a are in arithmetic progression. The three terms 16a, 2(b + c), and c are in geometric progression. The ratio of the common difference of the arithmetic progression to the common ratio of the geometric progression could be

-2
4
8
-16
0

Answer:

b+2c- (a+2b) =(c+2a)-(b+2c)

=>2b + 4c = c + 3a + 2b => a = c … (1)

(2b + 2c)² = 16ac=>4b² + 8bc + 4c² = 16ac …. (2)

From (1) and d(2) ,we get

4(b + c)² = 16c²=>b² + 2bc- 3c = 0

=>(b + 3c) (b- c) = 0=>b = c or – 3c

When b =c, the common difference

= (b + 2c)- (a + 2b) = 3c- 3c =0

the common ratio = \(\frac{2(b+c)}{16 a}=\frac{2(2 c)}{16 c}=\frac{1}{4}\)

their ratio = \(\frac{0}{1 / 4}=0\)

When b =-3c the common difference = (b + 2c)- (a + 2b) = (-3c + 2c)- (c- 6c) = 4c

the common ratio = \(\frac{2(b+c)}{16 a}=\frac{2(c-3 c)}{16 c}=\frac{-1}{4}\)

their ratio = \(\frac{4 c}{1 / 4}=-16 c\)

0 is given is the choices

Choice (5)

Chapter 6 Sequence And Series Step-By-Step Solutions

84. A number of saplings were lying at a point P, by the side of a straight road. Raju planted these saplings in a straight line with a distance of 20 meters between consecutive saplings. The first sapling was planted 20 m from P and all the saplings were planted on the same side of P. Raju carried only one sapling at a time and returned to P after planting each sapling, including the last one. If he travelled a total distance of 2.2 km, find the number of saplings that he planted.

Answer: To carry the first sapling to plant it and return, Raju has to travel 40 m. to carry the second one to plant it and return, Raju has to travel 80 m. Hence until he plants the final sapling (nth sapling) he will have to travel

40(1 + 2 + ….n) = 2200(given)

n(n+1)/2 = 55=>n² + n – 110 = 0=>n = 10 or- 11

Hence n = 10, as n > 0 Choice (1)

85. In an arithmetic progression, the sum of the squares of the seventh term and the eleventh term is 482. The product of its second term and the sixteenth term is 29. Find the product of its first and the seventeenth terms.

31
-31
-33
33
None of these

Answer: Let the first term and common difference of the arithmetic progression be a and d respectively.

Given that t_7² + t_11² = 482

=>2a² + 32ad + 136d² = 482 …. (1)

And (a + d) (a + 15d) = 29

=a² + 16ad+15d²=29………..(2)

By (1) – 2 x (2), we get

106d² = 482- 58 = 424

=>d² = 4………(3)

Consider, t₁ x t₁₇ = a(a + 16d) = a² + 16 ad

= (2)- 15 x (3) = 29 -15X4 = -31

Choice (2)

86. The sum of the squares of the first and fifth term of an arithmetic progression is 218. If the third term is less than the first term, and the sum of these two terms is 16, find the first term.

Answer: Let the first term of the arithmetic progression be a and the common difference be d.

Sum of the squares of the first and the fifth term

=a²+(a+4d)²=218

a²+4ad+8d²=109……………. (1)

a+(a+2d)=16

=>a+d=8=>a=8-d……………..(2)

Substituting a=8-d in (1) (8-d)²+4(8-d)d+8d²=104

5d²+16d-45=0 D=-5 or 1.8 As t₃<t₁,d<0,d=5

Substituting=-5 in (2). A=13. Hence, the first term of the arithmetic progression is 13

Choice (4)

87. The sum of the first n terms of an arithmetic progression is given by 3n²+4n. Find the 6th term of the arithmetic progression.

Answer: Let the first term and the common difference of the arithmetic progression be a and d respectively.

n/2(2a+(n-1)d)=3n²+4n=n(3n+4).

Comparing both sides d/2 = 3 (Whatever multiplies of n2)

In L.H.S. equals that in R.H.S.)

d=6. 2a-d/2=4, a=7.

The 6lh term of the arithmetic progression = a+5d=37.

Choice (3)

88. The sum of all the perfect squares from 50 to 2500 is

39825
40975
41895
44785
42785

Answer: The Sum of all the perfect squares from 50 to 2500

=the (sum of all perfect squares from 1 to 2500) – (Sum of all squares from 1 to 50).

Sum of the squares first n natural numbers

\(\frac{n(n+1)(2 n+1)}{6}\)

We have 50 perfect squares from 1 to 2500 whose sum is 1/6(50)(51)(101)

=25x17x101 =171700/4 = 42925…..(1)

Sum of all perfect squares from 1 to 7

= 1² + 2² + 3² + 4² + 5² + 6² + 7²

= 1/6(7)(8)(15) = 140……(2)

Required sum = 42925- 140 = 42785.

Choice(5)

89. a₁ = 1.5 and for n > l,a_n = 3a_n-1 + 4.Find a₃₀

3³⁰ -1
3²⁹(7/2)-2
3²⁹(7/2) + 2
3³⁰ +1
3³⁰(7/2) + 2

Answer:

a₁ = 1.5

a₂ = 3a₁ + 4

a₃ = 9a₁ + 4(1 + 3)

a₄ = 27a₁ + 4(1 + 3 + 3²) … and

a₃₀ = 3²⁹a₁ + 4 + 4(3) + 4(3²) + (3²⁸)

\(=3^{29}(3 / 2)+\frac{4\left(3^{29}-1\right)}{2}=2\left(\frac{7}{2}\right) 3^{29}-2\)

Choice (2)

90. How many terms, at the maximum, of the progression 2, 5, 8……can be considered, if the sum of the terms is to be less than 3000?

Answer: Let the maximum number of terms for which the sum of the terms of the A.P is less than 3000 be n.

The A.P has a first term of 2 and a common difference of 3. Sum of first n terms

\(\left.=\frac{n}{2}[2(2)+n-1) 3\right]<3000\)

n[1+3n] < 6000.

going by the choices only when n = 44,

we have the value ofthe expression n(3n + 1)being less than 6000 .

Hence n = 44.

Choice (2)

CA Foundation Maths Solutions For Chapter 6 Sequence And Series Introduction

CA Foundation Maths Solutions For Chapter 6 Progressions

CA Foundation Maths Solutions For Chapter 6 Geometric Progression

CA Foundation Maths Solutions For Chapter 6 Exercise – 2: Geometric Mean

CA Foundation Maths Solutions For Chapter 6

CA Foundation Maths Solutions For Chapter 6 Exercise – 3 (Mix- AP and GP)

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