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