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

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$2, 6, 18$

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(B) Let the three consecutive terms of the $G.P.$ be $\frac{a}{r}, a, ar$.Given that the product of the terms is $216$:$\frac{a}{r} 	imes a 	imes ar = 216$$a^3 = 216$$a = 6$Given that the sum of the products taken two at a time is $156$:$(\frac{a}{r} 	imes a) + (a 	imes ar) + (\frac{a}{r} 	imes ar) = 156$$\frac{a^2}{r} + a^2r + a^2 = 156$Substitute $a = 6$:$\frac{36}{r} + 36r + 36 = 156$$\frac{36}{r} + 36r = 120$Divide by $12$:$\frac{3}{r} + 3r = 10$$3r^2 - 10r + 3 = 0$$(3r - 1)(r - 3) = 0$So,$r = 3$ or $r = \frac{1}{3}$.If $r = 3$,the terms are $\frac{6}{3}, 6, 6 	imes 3$,which are $2, 6, 18$.If $r = \frac{1}{3}$,the terms are $\frac{6}{1/3}, 6, 6 	imes \frac{1}{3}$,which are $18, 6, 2$.Thus,the numbers are $2, 6, 18$.

If the product of three consecutive terms of a $G.P.$ is $216$ and the sum of their products taken two at a time is $156$,then the numbers are:

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