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 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 numbers.
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.
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