Question

Find three numbers in G.P. whose sum is 38 and their product is 1728.

Answer 1

Let the terms of the the given G.P. be $\frac{a}{r}\mathit{,}\mathit{}a\mathrm{and}\mathit{}ar.$
Then, product of the G.P. = 1728
$\Rightarrow$ a³ = 1728
$\Rightarrow$a = 12
Similarly, sum of the G.P. = 38
$\Rightarrow \frac{a}{r}+a+ar=38$
Substituting the value of a
$$\begin{gathered} \frac{12}{r}+12+12r=38 \\ \Rightarrow 12r^2+12r+12=38r \\ \Rightarrow 12r^2-26r+12=0 \\ \Rightarrow 2\left(6r^2-13r+6\right)=0 \\ \Rightarrow 6r^2-13r+6=0 \\ \Rightarrow \left(3r-2\right)\left(2r-3\right)=0 \\ \Rightarrow r=\frac{2}{3},\frac{3}{2} \end{gathered}$$
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.