Worksheet Solutions: Geometric Progression | Mathematics Class 10 ICSE PDF Download

Section A — Very Short Answer (1 mark each)

Q1. Find the common ratio of the sequence 5, 15, 45, ...
Solution (stepwise):
r = (2nd term)/(1st term) = 15/5 = 3.

Q2. Find the 4th term of the G.P. 3, 6, 12, ...
Solution:
a = 3, r = 6/3 = 2.
t4 = a × r^(4-1) = 3 × 2^3 = 3 × 8 = 24.

Q3. The second term of a G.P. is 12 and r = 1/3. Find the first term.
Solution:
ar = 12, r = 1/3 → a = 12 / (1/3) = 12 × 3 = 36.

Q4. Find the geometric mean between 4 and 25.
Solution:
G = sqrt(4 × 25) = sqrt(100) = 10.

Q5. If a = 81 and r = 1/3, find t5.
Solution:
t5 = a × r^(5-1) = 81 × (1/3)^4 = 81 × (1/81) = 1.

Section B — Short Answer (2–3 marks each)

Q6. Find the 8th term of the G.P. 2/3, 4/3, 8/3, ...
Solution (stepwise):
First term a = 2/3. r = (4/3) / (2/3) = 2.
t8 = a × r^(8-1) = (2/3) × 2^7 = (2/3) × 128 = 256/3.
So t8 = 256/3 (which is 85 1/3).

Q7. In the G.P. 5, 5/2, 5/4, ... which term equals 0.3125?
Solution:
a = 5, r = (5/2)/5 = 1/2. Let term be tn = 0.3125 = 5/16.
tn = a × r^(n-1) → 5 × (1/2)^(n-1) = 5/16 → (1/2)^(n-1) = 1/16.
1/16 = (1/2)^4 → n − 1 = 4 → n = 5.
So 0.3125 is the 5th term.

Q8. Find the sum of first 6 terms of the G.P. 3, −6, 12, −24, ...
Solution:
a = 3, r = −6/3 = −2, n = 6. Use Sn = a × (1 − r^n) / (1 − r).
r^6 = (−2)^6 = 64.
Sn = 3 × (1 − 64) / (1 − (−2)) = 3 × (−63) / 3 = −63.
Hence S6 = −63.

Q9. Find the sum to infinity of the G.P. 5 + 5/2 + 5/4 + 5/8 + ...
Solution:
a = 5, r = 1/2 and |r| < 1. S∞ = a / (1 − r) = 5 / (1 − 1/2) = 5 / (1/2) = 10.

Q10. Insert three geometric means between 2 and 162. (So make a G.P. with 5 terms: 2, G1, G2, G3, 162.)
Solution:
Let r be common ratio and total terms = 5, so 162 = 2 × r^(5-1) = 2 × r^4.
r^4 = 162 / 2 = 81 → r = 81^(1/4) = 3 (since 3^4 = 81).
So G1 = 2 × 3 = 6, G2 = 6 × 3 = 18, G3 = 18 × 3 = 54.
Inserted means: 6, 18, 54.

Section C — Long Answer (4–5 marks each)

Q11. The 4th term of a G.P. is 54 and the 7th term is 1458. Find the first term a, the common ratio r, and the sum of the first 8 terms.
Solution (stepwise):
Given t4 = a × r^3 = 54 ...(1)
Given t7 = a × r^6 = 1458 ...(2)
Divide (2) by (1): (a r^6) / (a r^3) = r^3 = 1458 / 54 = 27.
So r^3 = 27 → r = 3.
Substitute in (1): a × 3^3 = 54 → a × 27 = 54 → a = 54 / 27 = 2.
Now sum of first 8 terms, r > 1 so use Sn = a × (r^n − 1) / (r − 1):
S8 = 2 × (3^8 − 1) / (3 − 1). 3^8 = 6561.
S8 = 2 × (6561 − 1) / 2 = 6560.
Answers: a = 2, r = 3, S8 = 6560.

Q12. A sequence is 486, 162, 54, ... and continues until a term equals 2/3. 
Find (i) how many terms are there, and 
(ii) the total (sum) of all those terms.
Solution (stepwise):
This is a G.P. with a = 486 and r = 162/486 = 1/3. Let the number of terms be n, and last term l = 2/3.
Use tn = a × r^(n-1): 2/3 = 486 × (1/3)^(n-1).
Divide both sides by 486: (1/3)^(n-1) = (2/3) / 486 = 2 / (3 × 486) = 2 / 1458 = 1 / 729.
But (1/3)^6 = 1 / 729, so n − 1 = 6 → n = 7. (i) n = 7 terms.
Now sum S7 = a × (1 − r^7) / (1 − r): r^7 = (1/3)^7 = 1/2187.
S7 = 486 × (1 − 1/2187) / (1 − 1/3) = 486 × (2186/2187) / (2/3).
Compute stepwise: 486 × (2186/2187) × (3/2) = (486 × 3/2) × (2186/2187) = 729 × (2186/2187).
Since 729/2187 = 1/3, 729 × (2186/2187) = 729 − 729/2187 = 729 − 1/3 = 728 2/3.
So S7 = 2186/3 = 728 2/3. (ii) Total = 2186/3 (or 728 2/3).

Q13. The sum to infinity of a G.P. is 8 and its second term is 2. Find the first term, the common ratio, and the sum of the first 6 terms.
Solution (stepwise):
Given ar = 2 → a = 2 / r. Given S∞ = a / (1 − r) = 8 (with |r| < 1).
Substitute a: (2 / r) / (1 − r) = 8 → 2 = 8 r (1 − r).
Divide both sides by 2: 1 = 4 r (1 − r).
Expand: 1 = 4r − 4r^2 → rearrange: 4r^2 − 4r + 1 = 0.
This is (2r − 1)^2 = 0 → 2r − 1 = 0 → r = 1/2 (double root).
Then a = 2 / r = 2 / (1/2) = 4.
Sum of first 6 terms: use Sn = a × (1 − r^6) / (1 − r). r^6 = (1/2)^6 = 1/64.
S6 = 4 × (1 − 1/64) / (1 − 1/2) = 4 × (63/64) / (1/2) = 4 × (63/64) × 2 = 8 × (63/64) = 63/8.
So answers: a = 4, r = 1/2, S6 = 63/8 (which is 7 7/8).