Question

Find the common ratio of GP whose first term is $3$, the last is $3072$ and the sum of the series is $4095$

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

Answer: (C)

$4$

Since, $a=3,a^{r-1}=3072$ and $s_n=\frac{a(1-r^n)}{1-r}$

$4095=\frac{a+a{r}^{n-1}r}{1-r}$

$\Rightarrow 4095=\frac{3-3072r}{1-r}$

$\Rightarrow 4095=\frac{3(1-1024r)}{1-r}$

$\Rightarrow 1365=\frac{1-1024r}{1-r}$

$\Rightarrow 1365(1-r)=1-1024r$

$\Rightarrow 1365-1365r=1-1024r$

$\Rightarrow 341r=1364$

$\Rightarrow r=4$

Then common ratio of GP is$4$