Question

# 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.

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Solution

## Solution.Let the three digits be $a,ar$ and $a{r}^{2}$.Then, according to the question,$\begin{array}{l}100a+10ar+a{r}^{2}\dots \left(\mathrm{i}\right)\\ a+a{r}^{2}=2ar+1\\ a\left({r}^{2}-2r+1\right)=1\\ a\left(r-1{\right)}^{2}=1\dots \left(\mathrm{ii}\right)\\ a+ar=\frac{2}{3}\left(ar+a{r}^{2}\right)\\ ⇒3+3r=2r+2{r}^{2}\\ ⇒2{r}^{2}-r-3=0\\ ⇒\left(r+1\right)\left(2r-3\right)=0\\ \therefore r=-1,\frac{3}{2}\end{array}$For $r=-1$$\begin{array}{l}a=\frac{1}{\left(-1-1{\right)}^{2}}\\ =\frac{1}{4}\notin 1r\ne -1\end{array}$.For $r=\frac{3}{2}$$\begin{array}{l}a=\frac{1}{{\left(\frac{3}{2}-1\right)}^{2}}\\ =\frac{1}{4}\notin 1r\ne -1\end{array}$.From equation (i) the number is $400+10×4×\frac{3}{2}+4×\frac{9}{4}=469$Hence, the number is $469$.

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