Question

What is the ${10}^{th}$ term of the H.P. $\frac{1}{13},\frac{1}{26},\frac{1}{39},....$?

• A. $\frac{1}{130}$
• B. $\frac{1}{131}$
• C. $\frac{2}{130}$
• D. $\frac{1}{260}$

Answer 1

Answer: (A)

$\frac{1}{130}$

The reciprocal of A.P. is H.P.
A.P. is in the form of $13,26,39,...$
So, common difference in A.P. is $13$
So, ${10}^{th}$ term of the A.P. is
$a_n=a+(n-1)d$
$a_{10}=13+(10-1)13$
$=13+117$
$=130$
Therefore, ${10}^{th}$ term of HP is $\frac{1}{130}$.