Question

If for a G.P. $S_3=31$ and $S_6=3906$ then find a

• A. 2
• B. 3
• C. 1
• D. 4

Answer 1

The correct option is C 1

Given, $S_3=31\Rightarrow S_3=\frac{a(r^3-1)}{r-1}=31$ ...(1)
and $S_6=3906\Rightarrow S_6=\frac{a(r^6-1)}{r-1}=3906$ ...(2)
Dividing (2) by (1), we get
$\frac{r^6-1}{r^3-1}=126$
$\Rightarrow \frac{(r^3-1)(r^3+1)}{r^3-1}=126$
$\Rightarrow r^3+1=126$
$\Rightarrow r^3=125$
$\Rightarrow r=5$
Putting r=5 in (1), we get
$\frac{a(5^3-1)}{5-1}=31$
$\Rightarrow 31a=31$
$\Rightarrow a=1$
Hence a=1 and r=5.