If the sum of reciprocals of first 11 terms of an H.P. series is 110, find 6^th term of H.P.

The sum of first 11 terms of an A.P. $=\frac{11}{2}[2a+(11-1)d]=110$

(Because H.P. is reciprocal of A.P.)

$\begin{array}{l}\Rightarrow \frac{11}{2}[2a+(11-1)d]=110\\ \Rightarrow 2a+10d=\frac{110\times 2}{11}\\ \Rightarrow 2a+10d=20\\ \Rightarrow 2(a+5d)=20\\ \Rightarrow a+5d=\frac{20}{2}\\ \Rightarrow a+5d=10\end{array}$

6^th term of A.P. $=10$ then

6^th term of H.P. $=\frac{1}{10}$.