Question:
Find the sum of first 100 even natural numbers which are divisible by 5.
Solution:
The first few even natural numbers which are divisible by 5 are 10, 20, 30, 40, ...
This is an AP in which a = 10, d = (20 − 10) = 10 and n = 100
The sum of n terms of an AP is given by
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$=\left(\frac{100}{2}\right) \times[2 \times 10+(100-1) \times 10] \quad[\because a=10, d=10$ and $n=100]$
$=50 \times[20+990]=50 \times 1010=50500$
Hence, the sum of the first hundred even natural numbers which are divisible by 5 is 50500.
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