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Find the sum of the first 20 even natural numbers .
A. 400
B. 410
C. 420
D. 430

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Last updated date: 20th Apr 2024
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Answer
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Hint: Even natural numbers are numbers which we can divide from 2.
First even natural numbers are 2.
To find sum we can use formula for sum of n terms of A.P as given below:
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]=\dfrac{n}{2}(a+l)$
Where a is first term , d is common difference , n is number of terms and l is last term of A.P

Complete step by step solution:
So first 20 even natural numbers are$2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40$
Let Sum is S then we can write
$\text{S = }2+4+6+8+\cdot \cdot \cdot \cdot +34+36+38+40$
It is an arithmetic series because the difference between successive terms is equal.
$4-2=6-4=8-6=.....................=40-38=2$
In above series first term(a) is 2 , last term (l) is 40 and number of terms is 20
So we can write sum as
$\Rightarrow S=\dfrac{n}{2}(a+l)$
$\Rightarrow S=\dfrac{20}{2}(2+40)$
$\Rightarrow S=10\times 42$
$\Rightarrow S=420$
 Sum of the first 20 even natural numbers is 420.
Option (c) is correct.

Note: In this question we need to be careful to choose last term. First even natural numbers are 2. So ${{20}^{th}}$ even natural numbers are $2\times 20=40$ because if we don’t have last term we can’t apply summation of Arithmetic series having last term.