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

Let the three digits be a, ar and ar2.
$$100 a+10 a r+a r^{2} \ldots(\mathrm{i}) \\ a+a r^{2}=2 a r+1 or a\left(r^{2}-2 r+1\right)=1 or a(r-1)^{2}=1 \ldots.(ii)\\ a+a r=\frac{2}{3}\left(a r+a r^{2}\right)\\ \Rightarrow 3+3 r=2 r+2 r^{2} \Rightarrow 2 r^{2}-r-3=0 \Rightarrow(r+1)(2 r-3)=0\\ \therefore r=-1,3 / 2\\ \text { For r=-1, }a=\frac{1}{(r-1)^{2}}=\frac{1}{4} \notin 1 \quad r \neq-1\\ For r=3 / 2, a=\frac{1}{\left(\frac{3}{2}-1\right)^{2}}=4\ from (ii)\\ \text {From (i), number is } 400+10.4 \cdot \frac{3}{2}+4 \cdot \frac{9}{4}=469$$