Find three numbers in G.P. whose sum is 38 and their product is 1728.
Let the terms of the the given G.P. be $\frac{a}{r}, a$ and $a r$.
Then, product of the G.P. = 1728
= a3 = 1728
= a = 12
Similarly, sum of the G.P. = 38
$\Rightarrow \frac{a}{r}+a+a r=38$
Substituting the value of a
$\frac{12}{r}+12+12 r=38$
$\Rightarrow 12 r^{2}+12 r+12=38 r$
$\Rightarrow 12 r^{2}-26 r+12=0$
$\Rightarrow 2\left(6 r^{2}-13 r+6\right)=0$
$\Rightarrow 6 r^{2}-13 r+6=0$
$\Rightarrow(3 r-2)(2 r-3)=0$
$\Rightarrow r=\frac{2}{3}, \frac{3}{2}$
Hence, the G.P. for $a=12$ and $r=\frac{2}{3}$ is 18,12 and 8 .
And, the G.P. for $a=12$ and $r=\frac{3}{2}$ is 8,12 and 18 .
Hence, the three numbers are 8, 12 and 18.
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