Question

Find a three digit numberwhose consecutive digits form a GP. 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, then the resulting digits will form an AP.

Answer 1

Let the three digits be `a,ar,ar^(2)`, then according to hypothesis
`100a+10ar+ar^(2)-792=100ar^(2)+10ar+a`
` implies 99a(1-r^(2))=792`
` implies a(1+r)(1-r)=8 " " "......(i)"`
and `a,ar+2,ar^(2)` are in AP.
Then, `2(ar+2)=a+ar^(2)`
` implies a(r^(2)-2r+1)=4 implies a(r-1)^(2)=4" " "......(ii)"`
On dividing Eq.(i) by Eq. (ii), we get
`(r+1)/(r-1)=-2 impliesr=(1)/(3)`
From Eq. (ii),`a=9`
Thus, digits are `9,3,1` and so the required number is 931.