(a) \(\frac{-27}{2}\)
\(x\), 2\(x\) + 2, 3\(x\) + 3, are in G.P.
⇒ (2\(x\) + 2)2 = \(x\)(3\(x\) + 3)
⇒ 4x2 + 8\(x\) + 4 = 3x2 + 3\(x\)
⇒ x2 + 5\(x\) + 4 = 0
⇒ (\(x\) + 4) (\(x\) + 1) = 0
⇒ \(x\) = – 1 or – 4
Let the fourth term of the given G.P. be a. Then
r = \(\frac{a}{3x+3}\) = \(\frac{2x+2}{x}\)
⇒ a = \(\frac{(2x+2)(3x+3)}x{}\)
When \(x\) = – 4, Fourth term a = \(\frac{(-8+2)(-24+3)}{-4}\)
= \(\frac{-6\times-21}{-4}\) = \(\frac{-27}{2}\)
When \(x\) = – 1, a = \(\frac{(-2+2)(-3+3)}{-1}\) = 0.
∴ Fourth term = \(-\frac{27}{2}\).
