Question

A number consists of three digits in G.P. The sum of the right hand and left hand digits exceeds twice the middle digit by 1 and the sum of left hand and middle digits is two third of the sum of the middle and right hand digits. Find the sum of digits of number?

Answer 1

Let the three digits be $a,ar,a{r}^2$ then the number will be
$100a+10ar+a{r}^2\cdots \left(1\right)$
Given, $a+a{r}^2=2ar+1$
$\Rightarrow a(r^2-2r+1)=1$
$\Rightarrow a(r-1{)}^2=1\cdots (2)$
Also given, $a+ar=\frac{2}{3}(ar+a{r}^2)$
$\Rightarrow 3+3r=2r+2r^2$
$\Rightarrow 2r^2-r-3=0$
$\Rightarrow (r+1)(2r-3)=0$
$\therefore r=-1,3/2$
for $r=-1,a=\frac{1}{(r-1{)}^2}=\frac{1}{4}\notin 1$ $\therefore r\ne 1$
for $r=3/2,a=\frac{1}{(\frac{3}{2}-1{)}^2}=4$ {From(2)}
From $\left(1\right)$, number is $400+\mathrm{10.4.}\frac{3}{2}+4.\frac{9}{4}=469$