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 $G.P.$For three terms $a, b, c$ to be in $G.P.$,the condition is $b^2 = ac$.Therefore,$(2x + 2)^2 = x(3x + 3

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$-13.5$

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(D) Given that $x, 2x + 2, 3x + 3$ are in $G.P.$For three terms $a, b, c$ to be in $G.P.$,the condition is $b^2 = ac$.Therefore,$(2x + 2)^2 = x(3x + 3)$.Expanding the terms: $4(x + 1)^2 = 3x(x + 1)$.Case $1$: If $x + 1 = 0$,then $x = -1$. The sequence becomes $-1, 0, 0$,which is not a valid $G.P.$ as the common ratio is undefined.Case $2$: If $x + 1 
eq 0$,we can divide by $(x + 1)$: $4(x + 1) = 3x$.$4x + 4 = 3x \Rightarrow x = -4$.The terms are $x = -4$,$2(-4) + 2 = -6$,$3(-4) + 3 = -9$.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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