Question

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

Answer 1

Let the required three numbers of G.P. are $\frac{a}{r^{\prime }}a$ and ar. Then, their sum $=\frac{a}{r}+a+ar=38$
$\Rightarrow a\left(\frac{1+r+r^2}{r}\right)=38$ ....... (i)
product $=\frac{a}{r}\times a\times ar=1728$
$\Rightarrow a^3=(12{)}^3$
$\therefore a=12$ ........ (ii)
Substitute the value of a, in equation (i), we get
$\therefore 12\times \left(\frac{1+r+r^2}{r}\right)=38$
$\Rightarrow 6+6r+6r^2=19r$
$\Rightarrow 6r^2-13r+6=0$
$\Rightarrow (3r-2)(2r-3)=0\therefore r=\frac{2}{3}$ or $\frac{3}{2}$
Hence, the required numbers are 18, 12, 8 or 8, 12, 18
$\therefore$ Greatest number = 18.