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,

Solution:

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

Sum $=14$

$\Rightarrow a+a r+a r^{2}=14$

$\Rightarrow a\left(1+r+r^{2}\right)=14$     ...(1)

According to the question, a + 1, ar + 1 and ar2 − 1 are  in A.P.

$\therefore 2(a r+1)=a+1+a r^{2}-1$

$\Rightarrow 2 a r+2=a+a r^{2}$

$\Rightarrow 2 a r+2=14-a r \quad[$ From (i) $]$

$\Rightarrow 3 a r=12$

$\Rightarrow a=\frac{4}{r}$

Putting $a=\frac{4}{r}$ in (i)

$\Rightarrow \frac{4}{r}\left(1+r+r^{2}\right)=14$

$\Rightarrow 4 r^{2}-10 r+4=0$

$\Rightarrow 4 r^{2}-8 r-2 r+4=0$

$\Rightarrow(4 r-2)(r-2)=0$

$\Rightarrow r=\frac{1}{2}, 2$

Putting $r=\frac{1}{2}$ in (ii), we get $a=8$.

So, the G.P. is 8,4 and 2 .

Similarly putting r = 2 in (ii), we get a = 2.

So, the G.P is 2, 4 and 8.