Question

If the sum of reciprocals of first $11$ terms of an HP series is $110$, find the $6$th term of HP.

• A. $\frac{1}{10}$
• B. $\frac{2}{5}$
• C. $\frac{3}{10}$
• D. $\frac{4}{9}$

Answer 1

The correct option is A $\frac{1}{10}$

Now from the HP formulae, it is clear the reciprocals of first $11$ terms will make an AP.

The sum of first $11$ terms of an AP $=[2a+(11–1)d]\frac{11}{2}=110$

$\Rightarrow 2a+10d=20$

$\Rightarrow a+5d=10$

Now there are two variables, but $a+5d=T_6$ in an AP series.

And reciprocal of $6^{th}$ term of AP series will give the $6^{th}$ term of corresponding HP series.

So, the $6^{th}$ term of HP series is $\frac{1}{10}$