Three numbers are in G.P. such that their sum | vedclass

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Question

(a) Let three number of $G.P. $ $i.e.$, $a,\,ar,\,a{r^2}$

$a + ar + a{r^2} = 38$ and ${a^3}{r^3} = 1728$ ..$(i)$

$a(1 + r + {r^2}) = 38$ ..$(ii)

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(a) Let three number of $G.P. $ $i.e.$, $a,\,ar,\,a{r^2}$

$a + ar + a{r^2} = 38$ and ${a^3}{r^3} = 1728$ ..$(i)$

$a(1 + r + {r^2}) = 38$ ..$(ii)$

From equation $(i)$, $ar = 12$

==> $a = \frac{{12}}{r}$

From equation $(ii)$ $\frac{{12}}{r}(1 + r + {r^2}) = 38$

on solving $r = 2/3$ and $a = \frac{{12 	imes 3}}{2} = 18$

Required number $18, 12, 8$

Maximum number $i.e.$, $18.$

Three numbers are in $G.P.$ such that their sum is $38$ and their product is $1728$. The greatest number among them is

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