Question

Find the three numbers constituting a $G.P$. If it is know that the sum of the numbers is equal to $26$ and that when $1,6$ and $3$ are added to them respectively, the new numbers are obtained which from an $A.P$.

Answer 1

Let $a,ar,a{r}^2$ be in $G.P$., where
$a(1+r+r^2)=26$
Also $a+1,ar+6,a{r}^2+3$ are in $A.P.$
$\therefore 2(ar+6)=(a+1)+(a{r}^2+3)$
or $a(r^2-2r+1)=8$
Dividing $\left(1\right)$ and $\left(2\right)$ and simplifying, we get
$18(r^2+1)=60r$ or $3r^2-10r+3=0$
$\therefore$ $(r-3)(3r-1)=0$ $\therefore r=3,\frac{1}{3}.$
Putting in (1), $a=2$, $18$
$\therefore$ The number are $2,6,18$ or $18,6,2$