ML Aggarwal AP GP Exe-9.5 Class 10 ICSE Maths Solutions. We Provide Step by Step Solutions on questions to find Sum of n Terms of GP as council prescribe guideline for upcoming board exam. Visit official Website  CISCE  for detail information about ICSE Board Class-10.

Sum of nth Terms of GP

ML Aggarwal AP GP Exe-9.5 Class 10 ICSE Maths Solutions Ch-9

Board	ICSE
Subject	Maths
Class	10th
Chapter-9	Arithmetic and Geometric Progression
Writer / Book	Understanding
Topics	Find Sum of nth Terms of GP
Academic Session	2025-2026

How to Find Sum of nth Terms of GP

ML Aggarwal Arithmetic and Geometric Progression Exe-9.5 Class 10 ICSE Maths Solutions

Que-1:  Find the sum of:

(i) 20 terms of the series 2 + 6 + 18 + …
Ans:  First term a = 2,
Common ratio r = 6/2 = 3
Number of terms n = 20
So, S20 = a(rn – 1)/r – 1
= 2(320 – 1)/3 – 1
= 2(320 – 1)/2
= 320 – 1
Therefore, S20 = 320 – 1
(ii) 10 terms of series 1 + √3 + 3 + …
Ans:  First term a = 1,
Common ratio r = √3/1 = √3
Number of terms n = 10
So, S10 = a(rn – 1)/r – 1
= 1((√3)10 – 1)/ √3 – 1
Multiplying (√3 + 1) for both numerator and denominator we get,
= ((√310 – 1) (√3 + 1))/ ((√3 – 1) (√3 + 1)
= (35 – 1) (√3 + 1))/3 – 1 … [by rationalizing the denominator]
= ((243 – 1)( √3 + 1))/2
= 242(√3 + 1)/2
= 121(√3 + 1)
Therefore, S10 = 121(√3 + 1
(iii) 6 terms of the GP 1, -2/3, 4/9, …
Ans:  First term a = 1,
Common ratio r = -2/3 × 1= -2/3
Number of terms n = 6
So, S6 = a(rn – 1)/r – 1
= 1[1 – (-2/3)6]/(1 + (2/3))
= (3/5) (1 – (-26/36))
= (3/5) (1 – (64/729))
= (3/5) ((729 – 64)/729)
= 3/5 × (665/729)
= 133/243
(iv) 5 terms and n terms of the series 1 + 2/3 + 4/9 + …
Ans:  First term a = 1,
Common ratio r = 2/3 × 1= -/3
Number of terms n = 5
So, Sn = a(1 – rn)/1 – r
= 1[1 – (2/3)n]/(1 – 2/3)
Sn = 3[1 – (2/3)n]
Then, S5 = 3[1 – (2/3)5]
= 3[1 – (32/243)]
= 3((243 – 32)/243)
= 211/81

Que-2: Find the sum of the series 81 – 27 + 9 … – 1/27

Ans:  First term a = 81
r = -27/81
= -1/3
Last term l = -1/27
Sn = (a – lr)/(l – r)
= [81 + ((1/27) × (-1/3)]/[1 + (1/3)]
= [(81 – (1/81))]/(4/3)
= (6561 – 1)/[81 × (4/3)]
= (6560 × 3)/(81 × 4)
= 1640/27

Que-3: The nth term of a G.P. is 128 and the sum of its n terms is 255. If its common ratio is 2, then find its first term.

Ans:  The nth term of a G.P. Tn = 128
The sum of its n terms Sn = 255
Common ratio r = 2
We know that, Tn = arn – 1
128 = a2n – 1
a = 128/2n – 1 … [equation (i)]
Also we know that, Sn = a(rn – 1)/(r – 1)
255 = a(2n – 1)/(2 – 1)
By cross multiplication we get,
255 = a(2n – 1)
a = 255/(2n – 1) … [equation (ii)]
Now, consider both the equation(i) and equation (ii)
255/(2n – 1) = 128/(2n – 1)
By cross multiplication we get,
255 × 2n – 1 = 128(2n – 1)
255 × 2n – 1 = 128 × 2n – 128
(255 × 2n)/2 = 128 × 2n – 128
255 × 2n = 256 × 2n – 256
256 × 2n – 255 × 2n = 256
By simplification,
2n = 256
2n = 28
By comparing both LHS and RHS, we get,
Then, 128 = a27
128 = a × 128
a = 128/128
a = 1
Therefore, the value of a is 1.

Que-4: (i) How many terms of the G.P. 3, 32, 33, … are needed to give the sum 120?

Ans: Terms of the G.P. 3, 32, 33, …
Sum of the terms = 120
The first term a = 3
r = 32/3
= 9/3
= 3
We know that, Sn = a(rn – 1)/r – 1 = 120
3(3n – 1)/(3 – 1) = 120
3(3n – 1)/2 = 120
By cross multiplication we get,
3n – 1 = (120 ×2)/3
3n – 1 = 240/3
3n – 1 = 80
3n = 80 +1
3n = 81
3n = 34
Therefore, n = 4
(ii) How many terms of the G.P. 1, 4, 16, … must be taken to have their sum equal to 341?
Ans:  Terms of the G.P. 1, 4, 16, …
Sum of the terms = 341
The first term a = 1
r = 4/1
= 4
We know that, Sn = a(rn – 1)/r – 1 = 341
1(4n – 1)/(4 – 1) = 341
1(4n – 1)/3 = 341
By cross multiplication we get,
4n – 1 = (341 × 3)
4n – 1 = 1023
4n = 1023 + 1
4n = 1024
ML Aggarwal Solutions for Class 10 Maths Chapter 9 Image 5
4n = 45
Therefore, n = 5

Que-5: How many terms of the 2/9 – 1/3 + ½ + … will make the sum 55/72?

Ans: Terms of G.P. is 2/9 – 1/3 + ½ + …
Sum of the terms = 55/72
The first term a = 2/9
r = -1/3 ÷ 2/9 = (-1/3) × (9/2) = – 3/2
We know that, Sn = a(rn – 1)/r – 1 = 55/72
[(2/9)(1 – (-3/2)n)]/(1 + (3/2)) = 55/72
1 – (-3/2)n = (55/72) × (5/2) × (9/2)
(1 – (-1))n (3/2)n = 275/32
1 + 1(3/2)n = 275/32
(3/2)n = 275/32 – 1
(3/2)n = (275 – 32)/32
(3/2)n = 243/32
(3/2)n = (3/2)5
Therefore, n = 5

Que-6: The 2nd and 5th terms of a geometric series are -½ and sum 1/16, respectively. Find the sum of the series up to 8 terms.

Ans:   a2 = -½
a5 = 1/16
We know that, a2 = arn – 1
= ar2 – 1
a2 = ar = -½ … [equation (i)]
a5 = ar5 – 1
= ar4
a5 = ar4 = 1/16 … [equation (ii)]
Now, dividing equation (ii) by (i) we get,
r3 = 1/16 ÷ (-½)
= (1/16) × (-2)
= -1/8
r3 = (-1/2)3
So, r = -½
ar = -½
a × (-½) = -½
a = – ½ × (-2/1)
a = 1
Therefore, a = 1 and r = -½
Then, S8 = a(1 – rn)/(1 – r)
= 1[1 – (-1/2)8]/(1 + ½)
= [1 – (1/256)]/(3/2)
= (255/256) × (2/3)
= (510/768)
= 85/128

Que-7: The first term of G.P. is 27 and 8th term is 1/81. Find the sum of its first 10 terms.

Ans:  First term a = 27
8th term a8 = 1/81
Then, an = arn – 1
a8 = ar8 – 1 = 1/81
a8 = ar7 = 1/81
ar7 = 1/81
27r7 = 1/81
r7 = 1/(81 × 27)
r7 = 1/2187
r7 = 1/(37)
r = 1/3
So, S10 = a(1 – rn)/(1 – r)
= 27[1 – (1/3)10]/(1 – 1/3)
= 27[1 – (1/310)]/((3 – 1)/3)
= ((27 × 3)/2) [1 – 1/310]
= (81/2) [1 – 1/310]

Que-8: Find the first term of the G.P. whose common ratio is 3, last term is 486 and the sum of whose terms is 728.

Ans: Common ratio r = 3
Last term = 486
Sum of the terms = 728
We know that, Sn = a(rn – 1)/(r – 1)
= a(3n – 1)/(3 – 1) = 728
a(3n – 1)/2 = 728
a(3n – 1) = 728 × 2
a(3n – 1) = 1456 … [equation (i)]
Then, last term = arn – 1
486 = a × 3n – 1
486 = a(3n/3)
486 × 3 = a3n
1458 = a3n … [equation (ii)]
Consider equation (i), a(3n – 1) = 1456
a3n – a = 1456
Substitute the value of a3n in equation (i),
1458 – a = 1456
a = 1458 – 1456
a = 2
Therefore, the first term a is 2.

Que-9: In a G.P. the first term is 7, the last term is 448, and the sum is 889. Find the common ratio.

Ans:  First term a is = 7
Then, last term is = 448
Sum = 889
We know that, last term = arn – 1
7rn -1 = 448
rn – 1 = 448/7
rn – 1 = 64 … [equation (i)]
So, sum = a(rn – 1)/(r – 1) = 889
7(rn – 1)/(r – 1) = 889
(rn – 1)/(r – 1) = 889/7
(rn – 1)/(r – 1) = 127 … [equation (ii)]
Consider the equation (i),
rn/r = 64
rn = 64r
Now substitute the value of rn in equation (ii),
(64r – 1)/(r – 1) = 127
64r – 1 = 127r – 127
127r – 64r = -1 + 127
63r = 126
r = 126/63
r = 2
Therefore, common ratio = 2

Que-10: Find the third term of a G.P. whose common ratio is 3 and the sum of whose first seven terms is 2186.

Ans: Common ratio r = 3
S7 = 2186
We know that, Sn = a(rn – 1)/(r – 1)
S7 = a(3n – 1)/(3 – 1)
2186 = a(3n – 1)/2
By cross multiplication,
(2186 × 2) = a(37 – 1)
(4372) = a(2187 – 1)
4372 = a2186
a = 4372/2186
a = 2
Then, a3 = ar3 – 1
= ar2
= 2 × 32
= 2 × 9
a3 = 18

Que-11: If the first term of a G.P. is 5 and the sum of first three terms is 31/5, find the common ratio.

Ans: First term of a G.P. is a = 5
The sum of first three terms is S3 = 31/5
We know that, Sn = a(rn – 1)/(r – 1)
S3 = a(r3 – 1)/(r – 1)
31/5 = 5(r3 – 1)/(r – 1)
31/(5 × 5) = (r3 – 1)/(r – 1)
31/25 = (r3 – 1)/(r – 1)
(r – 1) (r2 + r + 1)/(r – 1) = 31/25
r2 + r + 1 = 31/25
By cross multiplication we get,
25(r2 + r + 1) = 31
25r2 + 25r + 25 = 31
Transposing 31 from right hand side to left hand side it becomes – 31,
25r2 + 25r + 25 – 31 =
25r2 + 25r – 6 = 0
25r2 + 30r – 5r – 6 = 0
5r(5r + 6) – 1(5r + 6) = 0
(5r – 1) (5r + 6) = 0
Take 5r – 1 = 0
r = 1/5
or 5r + 6 = 0
r = -6/5
Therefore, common ratio r = 1/5 or -6/5.

—  : End of ML Aggarwal AP GP Exe-9.5 Class 10 ICSE Maths Solutions. : –

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