Question

The consecutive digits of a three digit number are in $G.P$. If the middle digit be increased by $2$ then they form an $A.P$. If $792$ is subtracted from this number then we get the number consisting of same three digits but in reverse order. Find the number.

Answer 1

You must know that
$528=500+20+8$
$xyz=100x+10y+z$
Choose three number in $G.P$ to be $a,ar,a{r}^2$
By given condition,
$a\left(ar\right)\left(a{r}^2\right)-792=\left(a{r}^2\right)\left(ar\right)a$
Also $a,(ar+2),\left(a{r}^2\right)$ constitute an A.P.
$\therefore 2(ar+2)=a+a{r}^2$
or $a(r^2-2r+1)=4$
or $a(r-1{)}^2=4$
Again from $\left(1\right)$, we have
$100a+10ar+a{r}^2-792=100a{r}^2+10ar+a$
or $99(r^2-1)a=-792$
$\therefore (r-1)(r+1)a=-8$
Dividing $\left(3\right)$ by $\left(2\right)$, we get
$\frac{r+1}{r-1}=-2$ or $r=\frac{1}{3}$ $\therefore a=9$
Hence the numbers are $9,3,1$ or $931$
Also $931-792=139$ i.e reverse of $931$
Again $9,3+2,1$ i.e $9,5,1$ are in A.P