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.

Let the first three terms be \(\frac{a}{r},\,a,\,ar\)

\(\frac{a}{r} + a + ar = 26\)
⇒ \(a(\frac{1}{r} + 1 + r) = 26\)

\(\frac{a^2}{r^2} + a^2 + a^2r^2 = 364\)
⇒ \(a^2(\frac{1}{r^2} + 1 + r^2) = 364\)

Let \(r = 2\)
Then,
\(\frac{1}{r} + 1 + r = \frac{1}{2} + 1 + 2 = \frac{7}{2}\)
Hence,
\(a . \frac{7}{2} = 26\)
\(a = \frac{52}{7}\)
This does not satisfy the sum condition.

Try the GP: \(2, 6, 18,...\)
The sum of first three terms = 26
The sum of their squares = 364
So \(2, 6, 18\) are the first three terms of the GP.