Question

Let $A_n$ be the sum of the first $n$ terms of the geometric series $704+\frac{704}{2}+\frac{704}{4}+\frac{704}{8}+\dots$ and $B_n$ be the sum of the first $n$ terms of the geometric series $1984-\frac{1984}{2}+\frac{1984}{4}+\frac{1984}{8}+\dots$ 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)

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