Question

If $\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\ldots$ 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}+\ldots 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}+\ldots\right) $
$ =\frac{1}{4}\left[\left(\frac{1}{1}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{13}\right)+\ldots\right.$
$ \left.\ldots+\left(\frac{1}{4 n-3}-\frac{1}{4 n+1}\right)\right]$
$\Rightarrow \frac{1}{4}\left(1-\frac{1}{4 n+1}\right)=\frac{27}{109} $
$ \Rightarrow n=27$