Question

What is the sum of the first $9$ terms in this harmonic series $\frac{1}{8},\frac{1}{6},\frac{1}{4},\frac{1}{2}$?

Answer 1

We know the formula for sum of nth term in arithmetic progression.
The reciprocal of arithmetic progression is harmonic progression.
$a=82,d=-2$
$S_n=\frac{n}{2}[2a_1+(n-1\left)d\right]$
$S_9=\frac{9}{2}[2\times 8+(9-1)-2]$
$S_9=4.5[16-16]$
$S_9=4.5$
Sum of first 9 terms of HP = $\frac{1}{4.5}$