Question

If $\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\dots$ to $n$ terms $=\frac{27}{109}$, then $n=$

• A. 21
• B. 27
• C. 63
• D. 189

Answer 1

Answer: (B)

$\frac{1}{1\cdot 5}+\frac{1}{5\cdot 9}+\frac{1}{9\cdot 13}+\dots n\text{ terms }=\frac{27}{109}$
$=\frac{1}{4}\left(\frac{5-1}{1\cdot 5}+\frac{9-5}{5\cdot 9}+\frac{13-9}{9\cdot 13}+\dots \right)$
$=\frac{1}{4}[\left(\frac{1}{1}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{13}\right)+\dots$
$\dots +\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)]$
$\Rightarrow \frac{1}{4}\left(1-\frac{1}{4n+1}\right)=\frac{27}{109}$
$\Rightarrow n=27$