Find the sum of the first 10 terms of the harmonic sequence $\frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}, . . .$

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Solution

We know the formula for sum of $n^{t h}$ term in arithmetic progression.
The reciprocal of arithmetic progression is harmonic progression.
$a = 4, d = 2$
$S_{n} = \frac{n}{2} [2 a_{1} + (n - 1) d]$
$S_{10} = \frac{10}{2} [8 + (10 - 1) 2]$
$S_{10} = 5 [26]$
$S_{10} = 130$
Sum of first $10$ terms of HP = $\frac{1}{130}$

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