Question

How many terms of the $G.P$. $3, \frac{3}{2}, \frac{3}{4}, ...$ are needed to give the sum $\frac{3069}{512}$ ?

• A. $9$
• B. $12$
• C. $8$
• D. $10$

Answer 1

Answer: (D)

Let $n$ be the number of terms needed. Given that
$a=3, r= \frac{1}{2} $ and
$S_{n}= \frac{3069}{512}$
Now, $S_{n} =\frac{a\left(1-r^{n}\right)}{1-r} $
$ \Rightarrow \frac{3069}{512} $
$ = \frac{3\left(1-\frac{1}{2^{n}}\right)}{1-\frac{1}{2}} = 6\left(1-\frac{1}{2^{n}} \right)$
$ \Rightarrow \frac{3069}{3072} = 1 -\frac{1}{2^{n} }$
$\Rightarrow \frac {1}{2^{n}}= 1 -\frac{3069}{3072} $
$= \frac{1}{1024}$
or $2^{n}= 1024$
$ = 2^{10}$,
which gives $n= 10$.