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 sum of the digits of the number ?

Answer 1

We know that $528-500+20+8xyz=100x+10y+z$
Choose three numbers in $G.P.$ to be $a,ar,a{r}^2$
By given condition $a\left(ar\right)\left(a{r}^2\right)-792=\left(a{r}^2\right)\left(ar\right)a$ ...(1)
Also $\left(a\right),(ar+2),\left(a{r}^2\right)$ constitute an $A.P.$
$\therefore 2(ar+2)=a+a{r}^2\Rightarrow a(r^2-2r+1)=4\Rightarrow a(r-1{)}^2=4$ ...(2)
Again from (1), we have $100a+10ar+a{r}^2-792=100$

$\Rightarrow a{r}^2+10ar+a\Rightarrow 99(r^2-1)a+-792\Rightarrow (r-1)(r+1)a=-8$ ...(3)
Dividing (3) by (2), we get$\frac{r+1}{r-1}=-2\Rightarrow r=\frac{1}{3}\therefore a=9$
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 $A.P.$