Question

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

Answer 1

Let the three numbers be a, ar, $a{r}^2$ in G.P., where a is first term and r is the common ratio

Then

$a+ar+a{r}^2=38$

$a(1+r+r^2)=38$ ........(i)

and

$\left(a\right)\left(ar\right)(ar{)}^2=1728$

$a^3{r}^3=1728=4^3{3}^3=(12{)}^3$

$a^3=\frac{{12}^3}{r^3}\Rightarrow \frac{12}{r}=a$

Putting $a=\frac{12}{r}in\left(i\right)$

$\Rightarrow \frac{12}{r}(1+r+r^2)=38$

$\Rightarrow 12+12r+12r^2=38r$

$\Rightarrow 12r^2-26r+12=0$

$\Rightarrow 6r^2-13r+6=0$

$\Rightarrow 6r^2-9r-4r+6=0$

$\Rightarrow 3r(3r-3)-2(3r-3)=0$

$\Rightarrow (3r-2)(2r-3)=0$

$\Rightarrow r=\frac{3}{2},\frac{2}{3}$

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, 18