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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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