# Class 11 RD Sharma Solutions – Chapter 20 Geometric Progressions- Exercise 20.3 | Set 2

**Question 12. Find the sum:**

**Solution:**

Given summation can be written as,

S =

= (1 + 1/2 + 1/2

^{2}+ . . . . 10 terms) + (1/5^{2}+ 1/5^{3}+ 1/5^{4}+ . . . . 10 terms)=

=

Therefore, sum of the series is.

**Question 13. The fifth term of a G.P. is 81 whereas its second term is 24. Find the series and sum of its first eight terms.**

**Solution:**

We know nth term of G.P. is given by, a

_{n}= ar^{n-1}.According to the question, we have,

=> ar

^{4}= 81 . . . . (1)=> ar = 24 . . . . (2)

Dividing (2) by (1),

=> r

^{3}= 81/24=> r

^{3}= 27/8=> r = 3/2

Putting r = 3/2 in (2), we get,

=> a = (24)(2)/3

=> a = 16

We know sum of n terms of G.P. is given by, S

_{n}= a(r^{n}−1)/(r−1).So, S

_{8}= 16[(3/2)^{8}−1]/[(3/2)−1]= 16[3

^{8}− 2^{8}]/2^{7}= 6305/8

As a = 16 and r = 3/2, series would be

16, 24, 36, 54, . . . .

Also, the sum of first 8 terms of G.P. is 6305/8.

**Question 14. If S**_{1}, S_{2}, S_{3} be respectively the sum of n, 2n, 3n terms of a G.P., then prove that S^{2}_{1}+S^{2}_{2} = S_{1}(S_{2}+S_{3}).

_{1}, S

_{2}, S

_{3}be respectively the sum of n, 2n, 3n terms of a G.P., then prove that S

^{2}

_{1}+S

^{2}

_{2}= S

_{1}(S

_{2}+S

_{3}).

**Solution:**

We are given,

S

_{1}= Sum of n terms = a[1−r^{n}]/(1−r)S

_{2}= Sum of 2n terms = a[1−r^{2n}]/(1−r)S

_{3}= Sum of 3n terms = a[1−r^{3n}]/(1−r)We have,

L.H.S. = S

^{2}_{1}+S^{2}_{2}=

=

=

=

And R.H.S. = S

_{1}(S_{2}+S_{3})=

=

=

=

= L.H.S.

Hence, proved.

**Question 15. Show that the ratio of the sum of n terms of a G.P. to the sum of terms from (n+1)**^{th} to (2n)^{th} term is 1/r^{n}.

^{th}to (2n)

^{th}term is 1/r

^{n}.

**Solution:**

We know sum of n terms of a G.P. is given by, S

_{1}= a[1−r^{n}]/(1−r).And sum of terms from (n+1)

^{th}to (2n)^{th}term will be,S

_{2}= ar^{n}+ ar^{n+1}+ ar^{n+2}+ . . . . ar^{2n-1}= ar

^{n}[1−r^{n}]/(1−r)L.H.S. =

= 1/r

^{n}= R.H.S.

Hence, proved.

**Question 16. If a and b are the roots of x**^{2} – 3x + p = 0 and c, d are roots of x^{2} – 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p):(q – p) = 17:15.

^{2}– 3x + p = 0 and c, d are roots of x

^{2}– 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p):(q – p) = 17:15.

**Solution:**

We are given that a, b, c, d are in G.P. Let’s suppose the common ratio is r.

So, b=ar, c=ar

^{2}and d=ar^{3}Now a and b are the roots of x

^{2}– 3x + p = 0.Sum of roots = a + b = 3

=> a + ar = 3

=> a(1+r) = 3 ….. (1)

Product of roots = ab = p

=> a(ar) = p

=> a

^{2}r = p ….. (2)And c, d are the roots of x

^{2}− 12x + q = 0Sum of roots = c + d = 12

=> ar

^{2}+ ar^{3}= 12=> ar

^{2}(1+r) = 12 ….. (3)Product of roots = cd = q

=> ar

^{2}(ar^{3}) = q=> a

^{2}r^{5}= q ….. (4)Dividing equation (3) by (1), we get,

=> =

=> r

^{2}= 4=> r = ±2

When r=2, from (1), we get,=> a(1+2) = 3

=> a = 1

Putting a=1 and r=2 in (2),

=> p = (1)

^{2}(2) = 2From (4) we get,

q = (1)

^{2}(2)^{5}= 32Now L.H.S. = = = =

= R.H.S.

When r=−2, from (1), we get,=> a(1−2) = 3

=> a = −3

Putting a=−3 and r=−2 in (2),

p = (−3)

^{2}(−2) = −18From (4) we get,

q = (−3)

^{2}(−2)^{5}= −288Now L.H.S. = = = =

= R.H.S.

Hence, proved.

**Question 17. How many terms of the G.P. 3, 3/2, 3/4, … are needed to give the sum 3069/512?**

**Solution: **

Given G.P. has first term(a) = 3, common ratio(r) = (3/2)/3 = 1/2 and sum of terms(S

_{n}) = 3069/512.We know sum of n terms of a G.P. is given by S

_{n}= a(r^{n}–1)/(r–1).=> 3069/512 = 3[1–(1/2)

^{n}] / [1–(1/2)]=> 2(2

^{n}–1)/(2^{n}) = 1023/512=> 1023(2)

^{n}= 1024(2)^{n}– 1024=> 2

^{n}= 1024=> n = 10

Therefore, 10 terms of the G.P. should be taken together to make 3069/512.

**Question 18. A person has 2 parents, 4 grandparents, 8 great **grandparents**, and so on. Find the number **of **his ancestors during the ten generations preceding his own.**

**Solution:**

We have the sequence, 2,4,8, . . . . which forms a G.P. with first term(a) = 2 and common ratio(r) = 4/2 = 2.

We know sum of n terms of a G.P. is given by S

_{n}= a(r^{n}–1)/(r–1).Number of ancestors during the ten generations = Sum of first 10 terms of the G.P.

S

_{10}= 2(2^{10}–1)/(2–1)= 2(1023)

= 2046

Therefore, the number his ancestors during the ten generations preceding his own is 2046.

**Question 19. S**_{1}, S_{2}, S_{3}, . . . ., S_{n} are the sums of n terms of G.P.’s whose first term is 1 in each and common ratios are 1,2,3, . . . ., n respectively, then prove that S_{1} + S_{2} + 2S_{3} + 3S_{4} + . . . . + (n–1)S_{n} = **1**^{n} + 2^{n} + 3^{n} + . . . . + n^{n}.

_{1}, S

_{2}, S

_{3}, . . . ., S

_{n}are the sums of n terms of G.P.’s whose first term is 1 in each and common ratios are 1,2,3, . . . ., n respectively, then prove that S

_{1}+ S

_{2}+ 2S

_{3}+ 3S

_{4}+ . . . . + (n–1)S

_{n}=

^{n}+ 2

^{n}+ 3

^{n}+ . . . . + n

^{n}.

**Solution:**

S

_{1}, S_{2}, S_{3}, . . . ., S_{n}are the sums of n terms of G.P.’s whose first term is 1 in each and common ratios are 1,2,3, . . . ., n respectively.Here for S

_{1}, series will be 1,1,1,1, . . . up to n terms as first term(a) and common ratio(r) both are equal to 1.So, S

_{1}= 1+1+1+1+ . . . . up to n terms = nSo, L.H.S. = S

_{1}+ S_{2}+ 2S_{3}+ 3S_{4}+ . . . . + (n–1)S_{n}=

= n + (2

^{n}– 1) + (3^{n}– 1) + (4^{n}– 1) + . . . . (n^{n}– 1)= n + (2

^{n}+ 3^{n}+ 4^{n}+. . . . + n^{n}) – (1 + 1 +1 + 1 + . . . . (n–1) terms)= n + (2

^{n}+ 3^{n}+ 4^{n}+. . . . + n^{n}) – n + 1= 1 + 2

^{n}+ 3^{n}+ 4^{n}+. . . . + n^{n}= 1

^{n}+ 2^{n}+ 3^{n}+ 4^{n}+. . . . + n^{n}= R.H.S.

Hence, proved.

**Question 20. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of the terms occupying the odd places. Find the common ratio of the G.P. **

**Solution:**

As number of terms is even, let the number of terms of the G.P. be 2n.

According to the question,

Sum of all terms = 5 (Sum of the terms occupying the odd places)

=> a

_{1}+ a_{2}+ a_{3}+ . . . . + a_{n}= 5 (a_{1}+ a_{3}+ a_{5}+ . . . . a_{2n–1})=> a + ar + ar

^{2}+ . . . . + ar^{n–1}= 5 (a + ar^{2}+ ar^{4}+ . . . . + ar^{2n–2})=> a(1–r

^{2n})/(1–r) = 5a(1–r^{2n})/(1–r^{2})=> a/(1–r) = 5a/(1–r

^{2})=> a/(1–r) = 5a/[(1–r)(1+r)]

=> 5/(1+r) = 1

=> 1+r = 5

=> r = 4

Therefore, the common ratio of the G.P. is 4.

**Question 21. Let a**_{n} be the nth term of the G.P. of positive numbers. Let ** and ****, such that **α ≠ β**. Prove that the common ratio of the G.P. is **α/β.

_{n}be the nth term of the G.P. of positive numbers. Let

**Solution:**

We have,

=> a

_{2}+ a_{4 }+ a_{6}+ . . . . + a_{200}= α=> ar + ar

^{3}+ ar^{5}+ . . . . + ar^{199}= α=> ar(r

^{2(100)}–1)/(r–1) = α=> ar(r

^{200}–1)/(r–1) = α . . . . (1)Also we have,

=> a

_{1}+ a_{3}+ a_{5}+ . . . . + a_{199}= β=> a + ar

^{2}+ ar^{4}+ . . . . + ar^{198}= β=> a(r

^{2(100)}–1)/(r–1) = β=> a(r

^{200}–1)/(r–1) = β . . . . (2)Dividing (2) by (1), we get,

=> =

=> r =

Hence, proved.

**Question 22. Find the sum of 2n terms of the series whose every even term is ****‘a’**** times the term before it and every odd term is ‘c’ times the term before it, the first term being unity.**

**‘a’**

**Solution:**

Suppose we have the series, a

_{1},a_{2},a_{3}, . . . . a_{n}.According to the question, we have, a

_{1}= 1, a_{2}= a, a_{3}= ca, a_{4}= a^{2}c, a_{5}= a^{2}c^{2}and so on.Now, sum of 2n terms of the series = a

_{1}+ a_{2}+ a_{3}+ . . . . + a_{2n}S

_{2n}= 1 + a + ca + a^{2}c + a^{2}c^{2}+ . . . . 2n terms= (1+a) + ca(1+a) + a

^{2}c^{2}(1+a) + . . . . n termsNow this is a G.P. with first term(a) = (1+a) and common ratio(r) = ca. So, we get

S

_{2n}=

Therefore, sum of 2n terms of the series is.