# Class 11 RD Sharma Solutions – Chapter 20 Geometric Progressions- Exercise 20.3 | Set 1

**Question 1. Find the sum of the following geometric progressions:**

**(i) 2, 6, 18, … to 7 terms**

**Solution:**

Given G.P. has first term(a) = 2, common ratio(r) = 6/2 = 3 and number of terms(n) = 7

We know sum of n terms of a GP is given by S

_{n}= a(r^{n}–1)/(r–1).S

_{7}= 2(3^{7}–1)/(3–1)= 2(3

^{7}–1)/2= 2187–1

= 2186

Therefore, sum of 7 terms of the G.P. is 2186.

**(ii) 1, 3, 9, 27, … to 8 terms**

**Solution:**

Given G.P. has first term(a) = 1, common ratio(r) = 3 and number of terms(n) = 8

We know sum of n terms of a GP is given by S

_{n}= a(r^{n}–1)/(r–1).S

_{8}= 1(3^{8}–1)/(3–1)= 6560/2

= 3280

Therefore, sum of 8 terms of the G.P. is 3280.

**(iii) 1, –1/2, 1/4, –1/8, ….**

**Solution:**

Given G.P. has first term(a) = 1, common ratio(r) = –1/2 and number of terms(n) is infinite.

We know sum of n terms of an infinite GP is given by S = a/(1–r).

S = 1/[1 – (–1/2)]

= 1/(3/2)

= 2/3

Therefore, sum of infinite terms of the G.P. is 2/3.

**(iv) (a**^{2} – b^{2}), (a – b), (a–b)/(a+b), … to n terms

^{2}– b

^{2}), (a – b), (a–b)/(a+b), … to n terms

**Solution:**

Given G.P. has first term(a) = (a

^{2}– b^{2}), common ratio(r) = (a – b)/(a^{2}– b^{2}) = 1/(a+b) and number of terms is n.We know sum of n terms of an infinite GP is given by S

_{n}= a(r^{n}–1)/(r–1).S

_{n}==

=

Therefore, sum of n terms of the G.P. is.

**(v) 4, 2, 1, 1/2 … to 10 terms**

**Solution:**

Given G.P. has first term(a) = 4, common ratio(r) = 2/4 = 1/2 and number of terms(n) = 10.

We know sum of n terms of a GP is given by S

_{n}= a(r^{n}–1)/(r–1).S

_{10}==

=

=

Therefore, sum of 10 terms of the G.P. is.

**Question 2. Find the sum of the following geometric series:**

**(i) 0.15 + 0.015 + 0.0015 + … to 8 terms**

**Solution:**

Given G.P. has first term(a) = 0.15, common ratio(r) = 0.015/0.15 = 1/10 and number of terms(n) = 8.

We know sum of n terms of a GP is given by S

_{n}= a(r^{n}–1)/(r–1).S

_{8}== 0.15 (10/9) (1 – 1/10

^{8})= (1/6) (1 – 1/10

^{8})

Therefore, sum of 8 terms of the G.P. is (1/6) (1 – 1/10^{8}).

**(ii) √2 + 1/√2 + 1/2√2 + …. to 8 terms**

**Solution:**

Given G.P. has first term(a) = √2, common ratio(r) = (1/√2)/√2 = 1/2 and number of terms(n) = 8.

We know sum of n terms of a GP is given by S

_{n}= a(r^{n}–1)/(r–1).S

_{8}= √2[(1/2)^{8}–1]/[(1/2)–1]= √2(1–1/256)/(1/2)

= √2 (255/256) (2)

= (255√2)/128

Therefore, sum of 8 terms of the G.P. is (255√2)/128.

**(iii) 2/9 – 1/3 + 1/2 – 3/4 + … to 5 terms**

**Solution:**

Given G.P. has first term(a) = 2/9, common ratio(r) = (–1/3)/(2/9) = –3/2 and number of terms(n) = 5.

We know sum of n terms of a GP is given by S

_{n}= a(r^{n}–1)/(r–1).S

_{5}==

=

Therefore, sum of 5 terms of the G.P. is.

**(iv) (x + y) + (x**^{2} + xy + y^{2}) + (x^{3} + x^{2}y + xy^{2} + y^{3}) + …. to n terms

^{2}+ xy + y

^{2}) + (x

^{3}+ x

^{2}y + xy

^{2}+ y

^{3}) + …. to n terms

**Solution:**

Given series can be written as,

S

_{n}= (x + y) + (x^{2}+ xy + y^{2}) + (x^{3}+ x^{2}y + xy^{2}+ y^{3}) + . . . . to n terms(x – y) S

_{n}= (x + y) (x – y) + (x^{2}+ xy + y^{2}) (x – y) . . . to n terms(x – y) S

_{n}= x^{2}– y^{2 }+ x^{3}+ x^{2}y + xy^{2}– x^{2}y – xy^{2}– y^{3 }. . . to n terms(x – y) S

_{n}= (x^{2}+ x^{3}+ x^{4}+ . . . n terms) + (y^{2}+ y^{3}+ y^{4}+ . . . n terms)(x – y) S

_{n}= x^{2}[(x^{n}– 1)/(x – 1)] – y^{2}[(y^{n}– 1)/(y – 1)]S

_{n}= [x^{2}[(x^{n}– 1)/(x – 1)] – y^{2}[(y^{n}– 1)/(y – 1)]]/(x – y)

Therefore, sum of n terms of series is [x^{2}[(x^{n}– 1)/(x – 1)] – y^{2}[(y^{n}– 1)/(y – 1)]]/(x – y).

**(v) 3/5 + 4/5**^{2} + 3/5^{3} + 4/5^{4} + … to n terms

^{2}+ 3/5

^{3}+ 4/5

^{4}+ … to n terms

**Solution:**

Given series can be written as,

S

_{n}= (3/5 + 3/5^{3 }+ . . . to n terms) + (4/5^{2}+ 4/5^{4}+ . . . to n terms)=

=

=

Therefore, sum of n terms of series is.

**(vi)**

**Solution:**

Given G.P. has first term(a) = , common ratio(r) = = and number of terms is n.

We know sum of n terms of a GP is given by S

_{n}= a(r^{n}–1)/(r–1).S

_{n}==

= –a i[1–(1+i)

^{-n}]

Therefore, sum of n terms of G.P. is –a i[1–(1+i)^{-n}].

**(vii) 1, –a, a**^{2}, –a^{3}, . . . . to n terms (a ≠ 1)

^{2}, –a

^{3}, . . . . to n terms (a ≠ 1)

**Solution:**

Given G.P. has first term(a) = 1, common ratio(r) = –a and number of terms is n.

We know sum of n terms of a GP is given by S

_{n}= a(r^{n}–1)/(r–1).S

_{n}= [(–a)^{n}–1]/(–a–1)= [1–(–a)

^{n}]/(a+1)

Therefore, sum of n terms of G.P. is [1–(–a)^{n}]/(a+1).

**(viii) x**^{3} + x^{5} + x^{7} + . . . . n terms

^{3}+ x

^{5}+ x

^{7}+ . . . . n terms

**Solution:**

Given G.P. has first term(a) = x, common ratio(r) = x

^{5}/x^{3}= x^{2}and number of terms is n.We know sum of n terms of a GP is given by S

_{n}= a(r^{n}–1)/(r–1).= x

^{3}[x^{2n}–1]/[x^{2}–1]

Therefore, sum of n terms of G.P. is x^{3}[x^{2n}–1]/[x^{2}–1].

**(ix) √7 + √21 + 3√7 + . . . . n terms**

**Solution:**

Given G.P. has first term(a) = √7, common ratio(r) = √21/√7 = √3 and number of terms = n.

We know sum of n terms of a GP is given by S

_{n}= a(r^{n}–1)/(r–1).S

_{n}= √7[(√3)^{n}–1]/(√3–1)

Therefore, sum of n terms of G.P. is √7[(√3)^{n}–1]/(√3–1).

**Question 3. Evaluate the following:**

**(i) **

**Solution:**

Given summation can be written as,

S

_{11}= (2+3^{1}) + (2+3^{2}) + (2+3^{3}) + . . . . + (2+3^{11})= 2(11) + (3

^{1}+ 3^{2}+ 3^{3}+ . . . . 3^{11})= 2(11) + 3(3

^{11}–1)/(3–1)= 22 + 265719

= 265741

Therefore, value of the summation is 265741.

**(ii) **

**Solution:**

Given summation can be written as,

S

_{n}= (2+3^{0}) + (2^{2}+3^{1}) + (2^{3}+3^{2}) + . . . . + (2^{n}+3^{n-1})= (2

^{1}+ 2^{2}+ 2^{3}+ . . . . + 2^{n}) + (3^{0}+ 3^{1}+ 3^{2}+ . . . . + 3^{n-1})= 2(2

^{n}–1)/(2–1) + 3^{0}(3^{n}–1)/(3–1)= 2(2

^{n}–1) + (3^{n}–1)/2

Therefore, value of the summation is 2(2^{n}–1) + (3^{n}–1)/2.

**(iii)**

**Solution:**

Given summation can be written as,

S

_{10-2+1}= S_{9}= 4^{2}+ 4^{3}+ 4^{4}+ . . . . 4^{10}= 4

^{2}(4^{9}–1)/(4–1)= 16[4

^{9}–1]/3

Therefore, value of the summation is 16[4^{9}–1]/3.

**Question 4. Find the sum of the series:**

**(i) 5 + 55 + 555 + … to n terms**

**Solution:**

We have S

_{n}= 5 + 55 + 555 + ….. up to n terms.Multiplying and dividing by 9, we get

= [9+99+999+…to n terms]

= [(10–1)+(10

^{2}–1)+(10^{3}–1)…to n terms]= [(10+10

^{2}+10^{3}+….n terms) – (1+1+1+…..n terms)]=

=

Therefore, the sum of the series up to n terms is.

**(ii) 7 + 77 + 777 + … to n terms**

**Solution:**

We have S

_{n}= 7 + 77 + 777 + … to n terms.Multiplying and dividing by 9, we get,

= [9+99+999+…to n terms]

= [(10–1)+(10

^{2}–1)+(10^{3}–1)…to n terms]= [(10+10

^{2}+10^{3}+….n terms) – (1+1+1+…..n terms)]=

=

Therefore, the sum of the series up to n terms is.

**(iii) 9 + 99 + 999 + … to n terms**

**Solution:**

We have S

_{n}= 9 + 99 + 999 + … to n terms. It can be written as,= (10–1)+(10

^{2}–1)+(10^{3}–1)…to n terms= (10+10

^{2}+10^{3}+….n terms) – (1+1+1+…..n terms)=

Therefore, the sum of the series up to n terms is.

**(iv) 0.5 + 0.55 + 0.555 + … to n terms**

**Solution:**

We have S

_{n}= 0.5 + 0.55 + 0.555 + … to n terms. It can be written as,=

=

=

=

=

Therefore, the sum of the series up to n terms is.

**(v) 0.6 + 0.66 + 0.666 + … to n terms**

**Solution:**

We have S

_{n}= 0.6 + 0.66 + 0.666 + … to n terms. It can be written as,=

=

=

=

=

Therefore, the sum of the series up to n terms is.

**Question 5. How many terms of the G.P. 3, 3/2, 3/4, … be taken together to make 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 6. How many terms of the series 2 + 6 + 18 + …. must be taken to make the sum equal to 728?**

**Solution: **

Given G.P. has first term(a) = 2, common ratio(r) = 6/2 = 3 and sum of terms(S

_{n}) = 728.We know sum of n terms of a G.P. is given by S

_{n}= a(r^{n}–1)/(r–1).=> 728 = 2[3

^{n}–1]/[3–1]=> 3

^{n}–1 = 728=> 3

^{n}= 729=> n = 6

Therefore, 6 terms of the G.P. must be taken together to make the sum equal to 728.

**Question 7. How many terms of the sequence √3, 3, 3√3,… must be taken to make the sum 39+ 13√3 ?**

**Solution:**

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

_{n}) = 39+ 13√3.We know sum of n terms of a G.P. is given by S

_{n}= a(r^{n}–1)/(r–1).=> 39+13√3 = √3[3

^{n/2}–1]/(√3–1)=> (39+13√3)(√3–1) = √3(3

^{n/2}–1)=> 39√3–39+39–13√3 = 3

^{(n+1)/2}–√3=> 3

^{(n+1)/2 }= 27√3=> 3

^{n/2}√3 = 27√3=> 3

^{n/2 }= 27=> n/2 = 3

=> n = 6

Therefore, 6 terms of the G.P. must be taken to make the sum 39+ 13√3.

**Question 8. The sum of n terms of the G.P. 3, 6, 12, … is 381. Find the value of n.**

**Solution:**

Given G.P. has first term(a) = 3, common ratio(r) = 6/3 = 2 and sum of terms(S

_{n}) = 381.We know sum of n terms of a G.P. is given by S

_{n}= a(r^{n}–1)/(r–1).=> 381 = 3(2

^{n}–1)/(2–1)=> 2

^{n}– 1 = 127=> 2

^{n}= 128=> n = 7

Therefore, value of n is 7.

**Question 9. The common ratio of a G.P. is 3, and the last term is 486. If the sum of these terms be 728, find the first term.**

**Solution:**

Given G.P. has common ratio(r) = 3, last term(a

_{n}) = 486 and sum of terms(S_{n}) = 728.We know nth term of a G.P. is given by a

_{n}= ar^{n-1}.=> 486 = a(3)

^{n-1}=> 486 = a(3)

^{n}/3=> a(3)

^{n}= 1458 . . . . (1)We know sum of n terms of a G.P. is given by S

_{n}= a(r^{n}–1)/(r–1).=> 728 = a(3

^{n}–1)/(3–1)=> 1456 = a(3)

^{n}–aUsing (1) in the equation, we get,

=> a = 1458 – 1456 = 2

Therefore, first term of the G.P. is 2.

**Question 10. The ratio of the sum of the first three terms is to that of the first 6 terms of a G.P. is 125 : 152. Find the common ratio.**

**Solution:**

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

_{n}= a(r^{n}–1)/(r–1).According to the question, we have,

=>

=> (r

^{3}–1)/(r^{6}–1) = 125/152=> 1/(r

^{3}+1) = 125/152=> 125r

^{3}+ 125 = 152=> r

^{3}= 27/125=> r = 3/5

Therefore, the common ratio is 3/5.

**Question 11. The 4th and 7th term of a G.P. are 1/27 and 1/729 respectively. Find the sum of n terms of the G.P.**

**Solution:**

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

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

=> ar

^{3}= 1/27 . . . . (1)=> ar

^{6}= 1/729 . . . . (2)Dividing (2) by (1), we get,

=> r

^{3}= 27/729 = 1/27=> r = 1/3

Putting r = 1/3 in (1), we get,

=> a(1/3)

^{3}= 1/27=> a(1/27) = 1/27

=> a = 1

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

_{n}= a(r^{n}–1)/(r–1).=> S

_{n}= 1[(1/3)^{n}–1]/[(1/3)–1]= 3[1–(1/3)

^{n}]/2

Therefore, sum of n terms of the G.P. is 3[1–(1/3)^{n}]/2.

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