If $x, 2 x + 2$ and $3 x + 3$ are in GP, then its 4th term is

- 27

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- 13.5

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27

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13.5

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Solution

The correct option is C $- 13.5$
Given $x, 2 x + 2, 3 x + 3 ϵ$ G.P

$\Rightarrow (2 x + 2)^{2} = x (3 x + 3)$

$\Rightarrow 4 x^{2} + 8 x + 4 = 3 x^{2} + 3 x$

$\Rightarrow x^{2} + 5 x + 4 = 0$

$\Rightarrow x^{2} + x + 4 x + 4 = 0$

$\Rightarrow x = - 1 (not possible because second and third term will be equal to zero),$

$∴ x = - 4$

Therefore, G.P is $- 4, - 6, - 9. . . . . . . .$

Thus $T_{4} = (- 4) (\frac{- 6}{- 4})^{3} = - 13.5$

Hence, option 'B' is correct.

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