The product of three consecutive terms of a G | vedclass

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(D) Let the three consecutive terms of the $G.P.$ be $\frac{a}{r}, a, ar$.Given that the product of these terms is $512$:$\frac{a}{r} 	imes a 	imes

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$28$

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(D) Let the three consecutive terms of the $G.P.$ be $\frac{a}{r}, a, ar$.Given that the product of these terms is $512$:$\frac{a}{r} 	imes a 	imes ar = 512 \Rightarrow a^3 = 512 \Rightarrow a = 8$.According to the problem,if $4$ is added to the first and second terms,the new terms $\frac{8}{r} + 4, 8 + 4, 8r$ (i.e.,$\frac{8}{r} + 4, 12, 8r$) form an $A.P.$In an $A.P.$,the middle term is the arithmetic mean of the other two terms:$2 	imes 12 = (\frac{8}{r} + 4) + 8r$$24 = \frac{8}{r} + 4 + 8r$$20 = \frac{8}{r} + 8r$Divide by $4$: $5 = \frac{2}{r} + 2r$$2r^2 - 5r + 2 = 0$$(2r - 1)(r - 2) = 0$So,$r = 2$ or $r = \frac{1}{2}$.If $r = 2$,the terms are $\frac{8}{2}, 8, 8(2) = 4, 8, 16$.If $r = \frac{1}{2}$,the terms are $\frac{8}{1/2}, 8, 8(1/2) = 16, 8, 4$.In both cases,the sum of the terms is $4 + 8 + 16 = 28$.

The product of three consecutive terms of a $G.P.$ is $512$. If $4$ is added to each of the first and the second of these terms,the three terms now form an $A.P.$ Then the sum of the original three terms of the given $G.P.$ is

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