If x, 2x + 2,and 3x + 3 are in a geometric pr | vedclass

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(D) Given that $x, 2x + 2$,and $3x + 3$ are in a geometric progression $(GP)$.For a $GP$,the square of the middle term is equal to the product of the

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$-13.5$

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(D) Given that $x, 2x + 2$,and $3x + 3$ are in a geometric progression $(GP)$.For a $GP$,the square of the middle term is equal to the product of the first and third terms: $(2x + 2)^2 = x(3x + 3)$.Expanding this,we get $4x^2 + 8x + 4 = 3x^2 + 3x$.Rearranging the terms gives the quadratic equation $x^2 + 5x + 4 = 0$.Factoring the equation,we get $(x + 1)(x + 4) = 0$,which gives $x = -1$ or $x = -4$.If $x = -1$,the terms are $-1, 0, 0$,which cannot form a $GP$.If $x = -4$,the terms are $-4, -6, -9$. The common ratio $r = \frac{-6}{-4} = 1.5$.The fourth term $t_4 = a 	imes r^3 = -4 	imes (1.5)^3$.$t_4 = -4 	imes 3.375 = -13.5$.

If $x, 2x + 2$,and $3x + 3$ are in a geometric progression,what is the fourth term?

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