The product of three consecutive terms of a g | vedclass

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(B) Let the three terms in geometric progression be $\frac{a}{r}, a, ar$.The product of these terms is $(\frac{a}{r}) 	imes a 	imes (ar) = a^3 = 216

Vedclass · Accepted Answer

$2, 6, 18$

Vedclass · Answer

(B) Let the three terms in geometric progression be $\frac{a}{r}, a, ar$.The product of these terms is $(\frac{a}{r}) 	imes a 	imes (ar) = a^3 = 216 = (6)^3$,so $a = 6$.The sum of the products of these terms taken two at a time is $(\frac{a}{r} 	imes a) + (a 	imes ar) + (ar 	imes \frac{a}{r}) = 156$.Substituting $a = 6$: $\frac{36}{r} + 36r + 36 = 156$.$\frac{36}{r} + 36r = 120$.Dividing by $12$: $\frac{3}{r} + 3r = 10$.$3r^2 - 10r + 3 = 0$.$(3r - 1)(r - 3) = 0$.So,$r = 3$ or $r = \frac{1}{3}$.For $a = 6$ and $r = 3$,the terms are $\frac{6}{3}, 6, 6 	imes 3$,which are $2, 6, 18$.

The product of three consecutive terms of a geometric progression is $216$ and the sum of the products of these terms taken two at a time is $156$. Find the terms.

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