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Question

(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

Vedclass · Accepted Answer

$2, 6, 18$

Vedclass · Answer

(B) Let the three consecutive terms of the $G.P.$ be $\frac{a}{r}, a, ar$.Given that the product of the terms is $216$:$\frac{a}{r} 	imes a 	imes ar = 216 \Rightarrow a^3 = 216 \Rightarrow a = 6$.Given that the sum of the products taken two at a time is $156$:$\frac{a}{r} 	imes a + a 	imes ar + \frac{a}{r} 	imes ar = 156$.Substituting $a = 6$:$\frac{36}{r} + 36r + 36 = 156$.$\frac{36}{r} + 36r = 120$.Dividing by $12$,we get $\frac{3}{r} + 3r = 10$.$3 + 3r^2 = 10r \Rightarrow 3r^2 - 10r + 3 = 0$.$(3r - 1)(r - 3) = 0$.Thus,$r = 3$ 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 $18, 6, 2$.Therefore,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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