Find the sum of 100 term of the AP
Question:

Find the sum of 100 term of the AP $0.6,0.61,0.62,0.63, \ldots$

 

Solution:

To Find: The sum of 100 terms of the given AP series.

Sum of n terms of an AP with first term a and common difference d is given by

$S=\frac{n}{2}[2 a+(n-1) d]$

Here a = 0.6, n = 100, d = 0.01

$\Rightarrow S=\frac{100}{2}[1.2+99 \times 0.01]$

$=50[1.2+0.99]$

$=50 \times 2.19$

$109.5$

Sum of the series is 109.5

 

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