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

• Last Updated : 19 Apr, 2021

### (i) 2, 6, 18, â€¦ 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 Sn = a(rnâ€“1)/(râ€“1).

S7 = 2(37â€“1)/(3â€“1)

= 2(37â€“1)/2

= 2187â€“1

= 2186

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

### (ii) 1, 3, 9, 27, â€¦ 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 Sn = a(rnâ€“1)/(râ€“1).

S8 = 1(38â€“1)/(3â€“1)

= 6560/2

= 3280

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

### (iii) 1, â€“1/2, 1/4, â€“1/8, â€¦.

Solution:

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

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

S = 1/[1 â€“ (â€“1/2)]

= 1/(3/2)

= 2/3

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

### (iv) (a2 â€“ b2), (a â€“ b), (aâ€“b)/(a+b), â€¦ to n terms

Solution:

Given G.P. has first term(a) = (a2 â€“ b2), common ratio(r) = (a â€“ b)/(a2 â€“ b2) = 1/(a+b) and number of terms is n.

We know sum of n terms of an infinite GP is given by Sn = a(rnâ€“1)/(râ€“1).

Sn

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

### (v) 4, 2, 1, 1/2 â€¦ 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 Sn = a(rnâ€“1)/(râ€“1).

S10

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

### (i) 0.15 + 0.015 + 0.0015 + â€¦ 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 Sn = a(rnâ€“1)/(râ€“1).

S8

= 0.15 (10/9) (1 â€“ 1/108)

= (1/6) (1 â€“ 1/108)

Therefore, sum of 8 terms of the G.P. is (1/6) (1 â€“ 1/108).

### (ii) âˆš2 + 1/âˆš2 + 1/2âˆš2 + â€¦. to 8 terms

Solution:

Given G.P. has first term(a) = âˆš2, common ratio(r) = (1/âˆš2)/âˆš2 = 1/2 and number of terms(n) = 8.

We know sum of n terms of a GP is given by Sn = a(rnâ€“1)/(râ€“1).

S8 = âˆš2[(1/2)8â€“1]/[(1/2)â€“1]

= âˆš2(1â€“1/256)/(1/2)

= âˆš2 (255/256) (2)

= (255âˆš2)/128

Therefore, sum of 8 terms of the G.P. is (255âˆš2)/128.

### (iii) 2/9 â€“ 1/3 + 1/2 â€“ 3/4 + â€¦ to 5 terms

Solution:

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

We know sum of n terms of a GP is given by Sn = a(rnâ€“1)/(râ€“1).

S5

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

### (iv) (x + y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3) + â€¦. to n terms

Solution:

Given series can be written as,

Sn = (x + y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3) + . . . . to n terms

(x â€“ y) Sn = (x + y) (x â€“ y) + (x2 + xy + y2) (x â€“ y) . . . to n terms

(x â€“ y) Sn = x2 â€“ y2 + x3 + x2y + xy2 â€“ x2y â€“ xy2 â€“ y3 . . . to n terms

(x â€“ y) Sn = (x2 + x3 + x4 + . . . n terms) + (y2 + y3 + y4 + . . . n terms)

(x â€“ y) Sn = x2[(xn â€“ 1)/(x â€“ 1)] â€“ y2[(yn â€“ 1)/(y â€“ 1)]

Sn = [x2[(xn â€“ 1)/(x â€“ 1)] â€“ y2[(yn â€“ 1)/(y â€“ 1)]]/(x â€“ y)

Therefore, sum of n terms of series is [x2[(xn â€“ 1)/(x â€“ 1)] â€“ y2[(yn â€“ 1)/(y â€“ 1)]]/(x â€“ y).

### (v) 3/5 + 4/52 + 3/53 + 4/54 + â€¦ to n terms

Solution:

Given series can be written as,

Sn = (3/5 + 3/53 + . . . to n terms) + (4/52 + 4/54 +  . . . 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 Sn = a(rnâ€“1)/(râ€“1).

Sn

= â€“a i[1â€“(1+i)-n]

Therefore, sum of n terms of G.P. is â€“a i[1â€“(1+i)-n].

### (vii) 1, â€“a, a2, â€“a3, . . . . to n terms (a â‰  1)

Solution:

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

We know sum of n terms of a GP is given by Sn = a(rnâ€“1)/(râ€“1).

Sn = [(â€“a)nâ€“1]/(â€“aâ€“1)

= [1â€“(â€“a)n]/(a+1)

Therefore, sum of n terms of G.P. is [1â€“(â€“a)n]/(a+1).

### (viii) x3 + x5 + x7 + . . . . n terms

Solution:

Given G.P. has first term(a) = x, common ratio(r) = x5/x3 = x2 and number of terms is n.

We know sum of n terms of a GP is given by Sn = a(rnâ€“1)/(râ€“1).

= x3[x2nâ€“1]/[x2â€“1]

Therefore, sum of n terms of G.P. is x3[x2nâ€“1]/[x2â€“1].

### (ix) âˆš7 + âˆš21 + 3âˆš7 + . . . . n terms

Solution:

Given G.P. has first term(a) = âˆš7, common ratio(r) = âˆš21/âˆš7 = âˆš3 and number of terms = n.

We know sum of n terms of a GP is given by Sn = a(rnâ€“1)/(râ€“1).

Sn = âˆš7[(âˆš3)nâ€“1]/(âˆš3â€“1)

Therefore, sum of n terms of G.P. is âˆš7[(âˆš3)nâ€“1]/(âˆš3â€“1).

### Question 3. Evaluate the following:

(i)

Solution:

Given summation can be written as,

S11 = (2+31) + (2+32) + (2+33) + . . . . + (2+311)

= 2(11) + (31 + 32 + 33 + . . . . 311)

= 2(11) + 3(311â€“1)/(3â€“1)

= 22 + 265719

= 265741

Therefore, value of the summation is 265741.

(ii)

Solution:

Given summation can be written as,

Sn = (2+30) + (22+31) + (23+32) + . . . . + (2n+3n-1)

= (21 + 22 + 23 + . . . . + 2n) + (30 + 31 + 32 + . . . . + 3n-1)

= 2(2nâ€“1)/(2â€“1) + 30(3nâ€“1)/(3â€“1)

= 2(2nâ€“1) + (3nâ€“1)/2

Therefore, value of the summation is 2(2nâ€“1) + (3nâ€“1)/2.

(iii)

Solution:

Given summation can be written as,

S10-2+1 = S9 = 42 + 43 + 44 + . . . . 410

= 42(49â€“1)/(4â€“1)

= 16[49â€“1]/3

Therefore, value of the summation is 16[49â€“1]/3.

### (i) 5 + 55 + 555 + â€¦ to n terms

Solution:

We have Sn = 5 + 55 + 555 + â€¦.. up to n terms.

Multiplying and dividing by 9, we get

[9+99+999+â€¦to n terms]

[(10â€“1)+(102â€“1)+(103â€“1)â€¦to n terms]

[(10+102+103+â€¦.n terms) â€“ (1+1+1+â€¦..n terms)]

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

### (ii) 7 + 77 + 777 + â€¦ to n terms

Solution:

We have Sn = 7 + 77 + 777 + â€¦ to n terms.

Multiplying and dividing by 9, we get,

[9+99+999+â€¦to n terms]

[(10â€“1)+(102â€“1)+(103â€“1)â€¦to n terms]

[(10+102+103+â€¦.n terms) â€“ (1+1+1+â€¦..n terms)]

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

### (iii) 9 + 99 + 999 + â€¦ to n terms

Solution:

We have Sn = 9 + 99 + 999 + â€¦ to n terms. It can be written as,

= (10â€“1)+(102â€“1)+(103â€“1)â€¦to n terms

= (10+102+103+â€¦.n terms) â€“ (1+1+1+â€¦..n terms)

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

### (iv) 0.5 + 0.55 + 0.555 + â€¦ to n terms

Solution:

We have Sn = 0.5 + 0.55 + 0.555 + â€¦ 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 + â€¦ to n terms

Solution:

We have Sn = 0.6 + 0.66 + 0.666 + â€¦ 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, â€¦ 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(Sn) = 3069/512.

We know sum of n terms of a G.P. is given by Sn = a(rnâ€“1)/(râ€“1).

=> 3069/512 = 3[1â€“(1/2)n] / [1â€“(1/2)]

=> 2(2nâ€“1)/(2n) = 1023/512

=> 1023(2)n = 1024(2)n â€“ 1024

=> 2n = 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 + â€¦. 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(Sn) = 728.

We know sum of n terms of a G.P. is given by Sn = a(rnâ€“1)/(râ€“1).

=> 728 = 2[3nâ€“1]/[3â€“1]

=> 3nâ€“1 = 728

=> 3n = 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 âˆš3, 3, 3âˆš3,â€¦ must be taken to make the sum 39+ 13âˆš3 ?

Solution:

Given G.P. has first term(a) = 2, common ratio(r) = 3/âˆš3 = 1/âˆš3 and sum of terms(Sn) = 39+ 13âˆš3.

We know sum of n terms of a G.P. is given by Sn = a(rnâ€“1)/(râ€“1).

=> 39+13âˆš3 = âˆš3[3n/2â€“1]/(âˆš3â€“1)

=> (39+13âˆš3)(âˆš3â€“1) = âˆš3(3n/2â€“1)

=> 39âˆš3â€“39+39â€“13âˆš3 = 3(n+1)/2â€“âˆš3

=> 3(n+1)/2 = 27âˆš3

=>  3n/2 âˆš3 = 27âˆš3

=> 3n/2 = 27

=> n/2 = 3

=> n = 6

Therefore, 6 terms of the G.P. must be taken to make the sum 39+ 13âˆš3.

### Question 8. The sum of n terms of the G.P. 3, 6, 12, â€¦ 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(Sn) = 381.

We know sum of n terms of a G.P. is given by Sn = a(rnâ€“1)/(râ€“1).

=> 381 = 3(2nâ€“1)/(2â€“1)

=> 2n â€“ 1 = 127

=> 2n = 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(an) = 486 and sum of terms(Sn) = 728.

We know nth term of a G.P. is given by an = arn-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 Sn = a(rnâ€“1)/(râ€“1).

=> 728 = a(3nâ€“1)/(3â€“1)

=> 1456 = a(3)nâ€“a

Using (1) in the equation, we get,

=> a = 1458 â€“ 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 Sn = a(rnâ€“1)/(râ€“1).

According to the question, we have,

=>

=> (r3â€“1)/(r6â€“1) = 125/152

=> 1/(r3+1) = 125/152

=> 125r3 + 125 = 152

=> r3 = 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 an = arn-1.

According to the question, we have,

=> ar3 = 1/27   . . . . (1)

=> ar6 = 1/729 . . . . (2)

Dividing (2) by (1), we get,

=> r3 = 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 Sn = a(rnâ€“1)/(râ€“1).

=> Sn = 1[(1/3)nâ€“1]/[(1/3)â€“1]

= 3[1â€“(1/3)n]/2

Therefore, sum of n terms of the G.P. is 3[1â€“(1/3)n]/2.

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