If x, 2x + 2, 3x + 3 are in G.P.,then the fou | vedclass

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(D) Given that $x, 2x + 2, 3x + 3$ are in $G.P.$Therefore,$(2x + 2)^2 = x(3x + 3)$$4(x + 1)^2 = 3x(x + 1)$$4(x + 1)^2 - 3x(x + 1) = 0$$(x + 1)(4x + 4

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$-13.5$

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(D) Given that $x, 2x + 2, 3x + 3$ are in $G.P.$Therefore,$(2x + 2)^2 = x(3x + 3)$$4(x + 1)^2 = 3x(x + 1)$$4(x + 1)^2 - 3x(x + 1) = 0$$(x + 1)(4x + 4 - 3x) = 0$$(x + 1)(x + 4) = 0$So,$x = -1$ or $x = -4$.If $x = -1$,the terms are $-1, 0, 0$,which is not a valid $G.P.$ (as common ratio is undefined).If $x = -4$,the terms are $-4, -6, -9$.Here,the first term $a = -4$ and the common ratio $r = \frac{-6}{-4} = 1.5$.The fourth term $T_4 = ar^3 = (-4)(1.5)^3 = (-4)(3.375) = -13.5$.

If $x, 2x + 2, 3x + 3$ are in $G.P.$,then the fourth term is

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