If the sum of three terms of a G.P. is 19 and | vedclass

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(B) Let the three terms of the $G.P.$ be $\frac{a}{r}, a, ar$.Given that the product of the terms is $216$:$(\frac{a}{r}) \cdot a \cdot (ar) = 216$$a^

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$\frac{3}{2}$

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(B) Let the three terms of the $G.P.$ be $\frac{a}{r}, a, ar$.Given that the product of the terms is $216$:$(\frac{a}{r}) \cdot a \cdot (ar) = 216$$a^3 = 216$$a = 6$Given that the sum of the terms is $19$:$\frac{a}{r} + a + ar = 19$Substitute $a = 6$:$\frac{6}{r} + 6 + 6r = 19$$\frac{6}{r} + 6r = 13$Multiply by $r$:$6 + 6r^2 = 13r$$6r^2 - 13r + 6 = 0$Factorize the quadratic equation:$6r^2 - 9r - 4r + 6 = 0$$3r(2r - 3) - 2(2r - 3) = 0$$(3r - 2)(2r - 3) = 0$Thus,$r = \frac{2}{3}$ or $r = \frac{3}{2}$.

If the sum of three terms of a $G.P.$ is $19$ and their product is $216$,then the common ratio of the series is

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