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Question

(b) Let numbers are $\frac{a}{r},\;a,\;ar$

Under conditions, we get $\frac{a}{r}\;.\;a\;.\;ar = 216$

$ \Rightarrow $ $a = 6$

And sum of produ

Vedclass · Accepted Answer

$2, 6, 18$

Vedclass · Answer

(b) Let numbers are $\frac{a}{r},\;a,\;ar$

Under conditions, we get $\frac{a}{r}\;.\;a\;.\;ar = 216$

$ \Rightarrow $ $a = 6$

And sum of product pair wise $ = 156$

$ \Rightarrow $ $\frac{a}{r}\;.\;a + \frac{a}{r}\;.\;ar + a\;.\;ar = 156$

$ \Rightarrow $ $r = 3$

Hence numbers are $2, 6, 18.$

Trick : Since $2 	imes 6 	imes 18 = 216$ (as given) and no other option gives the value.

If the product of three consecutive terms of $G.P.$ is $216$ and the sum of product of pair-wise is $156$, then the numbers will be

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