If the product of three terms of a G.P. is 51 | vedclass

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(B) Let the three terms of a $G.P.$ be $\frac{a}{r}, a, ar$.Given that the product is $512$:$\frac{a}{r} 	imes a 	imes ar = 512$$a^3 = 512 = 8^3$$a

Vedclass · Accepted Answer

$4, 8, 16$

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(B) Let the three terms of a $G.P.$ be $\frac{a}{r}, a, ar$.Given that the product is $512$:$\frac{a}{r} 	imes a 	imes ar = 512$$a^3 = 512 = 8^3$$a = 8$.According to the second condition,$\frac{a}{r} + 8, a + 6, ar$ are in $A.P.$Substituting $a = 8$:$\frac{8}{r} + 8, 8 + 6, 8r$ are in $A.P.$$\frac{8}{r} + 8, 14, 8r$ are in $A.P.$For three terms $x, y, z$ to be in $A.P.$,$2y = x + z$:$2(14) = (\frac{8}{r} + 8) + 8r$$28 = \frac{8}{r} + 8 + 8r$$20 = \frac{8}{r} + 8r$Divide by $4$:$5 = \frac{2}{r} + 2r$$2r^2 - 5r + 2 = 0$$(2r - 1)(r - 2) = 0$$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 $\frac{8}{1/2}, 8, 8(1/2) = 16, 8, 4$.Both sets satisfy the condition. Thus,the numbers are $4, 8, 16$ or $16, 8, 4$.

If the product of three terms of a $G.P.$ is $512$. If $8$ is added to the first term and $6$ is added to the second term,the resulting numbers are in $A.P.$. Find the numbers.

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