In a HP 1-12-13-14-1100 is

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Standard XII

Mathematics

Method of Difference

In a HP 1+1...

Question

In a HP $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . . \frac{1}{100}$ is

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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.
$a = - 8, d = 3$
$S_{n} = \frac{n}{2} [a_{1} + a_{7}]$
$a_{n} = a + (n - 1) d$
$100 = 0.5 - 0.5 n$
$99.5 = - 0.5 n$
$n = 199$
$S_{199} = \frac{199}{2} [1 + 100]$
$S_{n} = 10049.5$
Sum of HP $= \frac{1}{10049.5}$

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In a HP 1+12+13+14+....1100 is

We know the formula for sum of nth term in arithmetic progression.The reciprocal of arithmetic progression is harmonic progression.a=−8,d=3Sn=n2[a1+a7]an=a+(n−1)d100=0.5−0.5n99.5=−0.5nn=199S199=1992[1+100]Sn=10049.5Sum of HP =110049.5

In a HP $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . . \frac{1}{100}$ is