Question

If the third and seventh terms of an H.P. are $\frac{1}{7}$ and $\frac{1}{15}$ respectively, then the $n^{th}$ term of the same H.P. is

• A. $2n+1$
• B. $\frac{1}{2n+1}$
• C. $2n-1$
• D. $\frac{1}{2n-1}$

Answer 1

The correct option is B $\frac{1}{2n+1}$

Let the first term of an H.P. be $\frac{1}{a}$ and the common difference of the denominators be $d$, then

$3^{rd}$ term of an H.P. is $\frac{1}{a+2d}=\frac{1}{7}$
$\Rightarrow a+2d=7$ ...(1)
Similarly, $7^{th}$ term is $\frac{1}{a+6d}=\frac{1}{15}$
$a+6d=15$ ...(2)
Subtracting (1) from (2), we get

$a=3,d=2$
$\therefore$ The $n^{th}$ of this H.P. is

$\frac{1}{a+(n-1)d}=\frac{1}{3+(n-1)\times 2}=\frac{1}{2n+1}$