Using a(1 + r + r²) = 26 and a²(1 + r² + r⁴) = 364, dividing squares gives (1 + r + r²) / (1 – r + r²) = 13/7, leading to r = 3 or 1/3. The terms are 2, 6, 18 or 18, 6, 2.
The sum of the first three terms of a geometric progression is 26 and the sum of their squares is 364. Find the terms of the GP.
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Let the terms be a, ar and ar².
Given:
a(1 + r + r²) = 26
Squaring: a²(1 + r + r²)² = 676
Sum of squares:
a²(1 + r² + r⁴) = 364
Using identity (1 + r² + r⁴) = (1 + r + r²)(1 – r + r²):
Dividing gives (1 + r + r²) / (1 – r + r²) = 676 / 364 = 13 / 7.
7 + 7r + 7r² = 13 – 13r + 13r²
6r² – 20r + 6 = 0, so 3r² – 10r + 3 = 0.
Thus r = 3 or r = 1/3.
If r = 3, a = 2, terms are 2, 6, 18.
If r = 1/3, a = 18, terms are 18, 6, 2.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-8/