Question

The sum of three numbers in G.P. is 14. If the first two terms are each increased by 1 and the third term decreased by 1, the resulting numbers are in A.P. Find the largest of those numbers $-$ smallest of those numbers ?

Answer 1

Let the numbers be $a,ar,a{r}^2$

$\therefore Sum=a(1+r+r^2)=14...\left(1\right)$ Also

$a+1,ar+1,4a{r}^2-1$ are in A.P

$\Rightarrow a(r^2-2r+1)=2..\left(2\right)$

Divide (1) and (2) and simplifying

$2r^2-5r+2=0\therefore r=2,\frac{1}{2}$

and hence $a=2,8$

$\therefore$ Numbers are $2,4,8$ or $8,4,2$

The required difference of $=6$