Question

Find a three digit number whose consecutive numbers form a G.P. If we subtract 792 from this number , we get a number consisting of the same digits written in the reverse order. Now if we increase the second digit of the required number by 2, the resulting number will form an A.P.

Answer 1

Let the three digits be $a,ar,a{r}^2$ then according to given condition,
$100a+10ar+a{r}^2+792=100a{r}^2+10ar+a$
$\Rightarrow a(r^2-1)=8\cdots \left(1\right)$
also $a,ar+2,a{r}^2$ are in A.P.
then $2(ar+2)=a+a{r}^2$
$\Rightarrow a(r^2-2r+1)=4\cdots \left(2\right)$
Dividing $\left(1\right)$ by $\left(2\right)$,
then $\frac{a(r^2-1)}{a(r^2-2r+1)}=\frac{8}{4}$
$\Rightarrow \frac{(r+1)(r-1)}{(r-1{)}^2}=2$
$\Rightarrow \frac{r+1}{r-1}=2$
$\therefore r=3$ from $\left(1\right)$, $a=1$
Thus, the digits are $1,3,9$ and the required number is $931$