A number consists of three digits which are in GP the sum of the right hand and left hand digits exceeds twice the middle digits by 1 and the sum of the left hand and middle digits is two thirds of the sum of the middle and right hand digits. Find the number.

Explanation

Let the three digits be a, ar and `ar^(2)` ,then number is
`100a+10ar+ar^(2) " " "……(i)"`
Given, `a+ar^(2)=2ar+1`
or `a(r^(2)-2r+1)=1`
or `a(r-1)^(2)=1" " "…….(ii)"`
Also, given `a+ar=(2)/(3)(ar+ar^(2)) implies 3+3r=2r+2r^(2)`
or `2r^(2)-r-3=0` or `(r+1)(2r-3)=0`
`:.r=-1,(3)/(2)`
For `r=-1, a=(1)/((r-1))^(2)=(1)/(4)cancel in I`
`:. rne-1`
For `r=(3)/(2),a=(1)/((3)/(2)-1)^(2)=4" " [" from Eq. (ii) " ]`
From eq.(i), number is `400+10*4*(3)/(2)+4*(9)/(2)=469`