If $\frac{1}{1·5}+\frac{1}{5·9}+\frac{1}{9·13}+\dots$ to $n$ terms $=\frac{27}{109}$ then $n=$

Given $\frac{1}{1·5}+\frac{1}{5·9}+\frac{1}{9·13}+\dots$ to $n$ terms $=\frac{27}{109}.$

We know that the $n^{th}$ term of an $A.P., a, a+d, a+2d,...$ is $a+\left(n-1\right)d.$

Hence, the $n^{th}$ term of $1, 5, 9,...$ is $1+\left(n-1\right)4=4n-3$ and that of $5, 9, 13,...$ is $5+\left(n-1\right)4=4n+1.$

Thus, the given series can be written as $\frac{1}{1·5}+\frac{1}{5·9}+\frac{1}{9·13}+\dots +\frac{1}{\left(4n-3\right)\left(4n+1\right)}=\frac{27}{109}$

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

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

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

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

$\Rightarrow \frac{1}{4}\left[1-\frac{1}{\left(4n+1\right)}\right]=\frac{27}{109}$

$\Rightarrow \frac{1}{4}\left[\frac{4n+1-1}{\left(4n+1\right)}\right]=\frac{27}{109}$

$\Rightarrow \frac{n}{4n+1}=\frac{27}{109}$

$\Rightarrow 109n=27\left(4n+1\right)$

$\Rightarrow 109n=108n+27$

$\Rightarrow n=27$