Question

If for a G.P. $S_6=126$ and $S_3=14$ then find r.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

Given, $S_3=31\Rightarrow S_3=\frac{a(r^3-1)}{r-1}=14$ ...(1)
and $S_6=3906\Rightarrow S_6=\frac{a(r^6-1)}{r-1}=126$ ...(2)
Dividing (2) by (1), we get
$\frac{r^6-1}{r^3-1}=9$
$\Rightarrow \frac{(r^3-1)(r^3+1)}{r^3-1}=9$
$\Rightarrow r^3+1=9$
$\Rightarrow r^3=8$
$\Rightarrow r=2$
Putting r=2 in (1), we get
$\frac{a(2^3-1)}{2-1}=14$
$\Rightarrow 7a=14$
$\Rightarrow a=2$
Hence a=2 and r=2.