If the sum of three terms of a Geometric Prog | vedclass

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(A) Let the three terms of the $GP$ be $\frac{a}{r}, a, ar$.The product of these terms is $\frac{a}{r} 	imes a 	imes ar = a^3 = 216$.Thus,$a = \sqrt

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$3/2$

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(A) Let the three terms of the $GP$ be $\frac{a}{r}, a, ar$.The product of these terms is $\frac{a}{r} 	imes a 	imes ar = a^3 = 216$.Thus,$a = \sqrt[3]{216} = 6$.The sum of the three terms is $\frac{6}{r} + 6 + 6r = 19$.Subtracting $6$ from both sides,we get $\frac{6}{r} + 6r = 13$.Multiplying by $r$,we get $6 + 6r^2 = 13r$,which simplifies to the quadratic equation $6r^2 - 13r + 6 = 0$.Factoring the quadratic equation: $6r^2 - 9r - 4r + 6 = 0 \implies 3r(2r - 3) - 2(2r - 3) = 0$.This gives $(3r - 2)(2r - 3) = 0$.Therefore,$r = \frac{2}{3}$ or $r = \frac{3}{2}$.

If the sum of three terms of a Geometric Progression $(GP)$ is $19$ and their product is $216$,then the common ratio of this $GP$ is:

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