Find the sum of the first 20 even natural numbers .
A. 400
B. 410
C. 420
D. 430
Answer
618.3k+ views
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.
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.
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