Question

What is the sum of the following harmonic series: $\frac{1}{-64},\frac{1}{-66},\frac{1}{-68},.....,\frac{1}{-100}$?

Answer 1

We know the formula for sum of $n^{th}$ term in arithmetic progression.
The reciprocal of arithmetic progression is harmonic progression.
$a=-64,d=-2$
$S_n=\frac{n}{2}[2a_1+(n-1\left)d\right]$
$a_n=a+(n-1)d$
$-100=-64+(n-1)(-2)$
Therefore, $n=19$
$S_{19}=\frac{19}{2}[2\times -64+(19-1\left)\right(-2\left)\right]$
$S_{19}=9.5[-128-36]$
$S_{19}=9.5[-164]$
$S_{19}=-1558$
Sum of HP $=\frac{-1}{1558}$