Question

Let $A_{n}$ be the sum of the first $n$ terms of the geometric series $704+\frac{704}{2}+\frac{704}{4}+\frac{704}{8}+\ldots$ and $B_{n}$ be the sum of the first $n$ terms of the geometric series $1984-\frac{1984}{2}+\frac{1984}{4}+\frac{1984}{8}+\ldots$ If $A_{n}=B_{n}$, then the value of $n$ is (where $n \in N$ ).

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

Answer: (B)

$A_{n}=704+\frac{704}{2}+\frac{704}{4}+\ldots$ to $n$ terms
$=\frac{704\left(1-\left(\frac{1}{2}\right)^{n}\right)}{1-\frac{1}{2}}=704 \times 2\left(1-\left(\frac{1}{2}\right)^{n}\right)$
$B_{n}=1984-\frac{1984}{2}+\frac{1984}{4} \ldots$ to $n$ terms
$=\frac{1984\left(1-\left(\frac{-1}{2}\right)^{n}\right)}{1-\left(\frac{-1}{2}\right)}=1984 \times \frac{2}{3}\left(1-\left(\frac{-1}{2}\right)^{n}\right)$
Now, $A_{n}=B_{n}$
$ \Rightarrow 704 \times 2\left(1-\left(\frac{1}{2}\right)^{n}\right)$
$=1984 \times \frac{2}{3} \times\left(1-\left(\frac{-1}{2}\right)^{n}\right)$
$\Rightarrow 33-31=33\left(\frac{1}{2}\right)^{n}-31\left(\frac{-1}{2}\right)^{n}$
$\Rightarrow 2^{n+1}=33-31(-1)^{n}$
$ \Rightarrow n=5$