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Find the sum of all even natural numbers from 2 to 40

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Last updated date: 28th Feb 2024
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IVSAT 2024
Answer
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Hint: To solve this question, firstly we will find the number of even number from 2 to 40 by using formula of finding ${{n}^{th}}$ term of an A.P when last term is given. After finding the value of n, we will find the summation of those n terms by taking a = 2 and l = 40, where a is the first term of an A.P and l is last term of an A.P.

Complete step-by-step answer:
Before solving this question, we must know about Natural Numbers and Even Numbers.
Natural numbers are those that are used for counting and ordering.
Example: 1, 2, 3, 4, 5……….100, 117, 674, 123, etc.
Any integer that can be divided exactly by 2 is an even number. The last digit of even numbers is 0, 2, 4, 6 or 8, i.e. the digits ending with 0, 2, 4, 6 or 8 are always divisible by 2 and thus, are even numbers.
Example: 26, 18, 20456, 2716, 2756, 204, etc. are all even numbers because they are divisible by 2.
To solve this question we will simply add the first natural numbers that are even from 2 to 40.
Let us now solve the question.
The even natural numbers from 2 to 40 are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38 and 40.
We know that, ${{n}^{th}}$ term of an A.P when last term is given can be find by formula \[l=a+(n-1)d\], where \[l={{a}_{n}}\]
As, l = 40 and a = 2
So, \[40=2+(n-1)2\]
On simplifying, we get
\[38=(n-1)2\]
\[(n-1)=19\]
n = 20
so, there are a total of 20 even numbers from 2 to 40.
Now, we know that the sum of terms of an arithmetic progression which as follows
\[{{S}_{n}}=\dfrac{n}{2}\left( a+l \right)\] , where \[l=a+(n-1)d\].
So, putting values of l, a, n and d, we get
\[{{S}_{20}}=\dfrac{20}{2}\left( 2+40 \right)\]
On simplifying, we get
\[{{S}_{20}}=10\times 42\]
Or, \[{{S}_{20}}=420\]
Hence, the sum of all even natural numbers from 2 to 40 is equals to 420.

Note: One must always know that ${{n}^{th}}$ term of an A.P is given can be find by formula \[{{a}_{n}}=a+(n-1)d\], also, remember that the sum of terms of an arithmetic progression which as follow\[{{S}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)\], where a = first term of an A.P, n = total number of terms of an A.P and d=common difference between consecutive terms in an A.P. Try not to make any calculation mistake.

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