# What is the sum of the terms of the series 2 , 6 , 10 , .. , 402 ?

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S = 2 + 6 + 10 + ..... + 402

Let 2, 6, 10, ..., 402 be a series such that:

s1= 2*1 = 2

s2= 2*3 = 6

s3 = 2*5 = 10

.....

sn = 2*(2n-1) = 402

==> 4n - 2 = 402

==> 4n = 404

==> n = 101

We know that:

Sn = (s1+sn)*(n/2)

S101 = (2+ 402)*(101)/2 = 20402

**==> the sum S = 20402**

In the given series we see that the difference between consecutive terms is 10-6 = 6-2 = 2

Therefore it is an AP. The first term is 2.

Now the nth term of an AP is given as a +(n-1)d

402= 2 + (n-1)4

=> 400 = (n-1)4

=> n-1 = 400/ 4

=> n=100 +1

=> n= 101

The sum of n terms of an AP is (a1+ an)*(n/2)

Here a1 = 2, an = 402, n= 101

So the sum is ( 2 + 402)*(101/2)

= 202 * 101

= 20402

**The required sum of the terms is 20402**

To find the sum of the series whose starting term = a1 = 2.

common diffrence d = 6-2 = 10-6 =4.

The last nth term an = 402.

Therefore the number of terms n = (an-a1)/d +1 = (402 -2)/4 +1 = 101.

Theherefore sum Sn = (a1+an)*n/2 = (2+402)(101)/2 = 20402

We notice that if we'll calculate the difference between 2 consecutive terms of the given series, we'll obtain the same value each time:

So, the given series is an arithmetic progression whose common difference is d = 4.

We can calculate the sum of n terms of an arithmetic progression in this way;

Sn = (a1 + an)*n/2

a1 - the first term of the progression

a1 = 2

an - the n-th term of the progression

an = 402

n - the number of terms

We can notice that we know the first and the last terms but we don't know the number of terms. We can calculate the number of terms using the formula of general term.

an = a1 + (n-1)*d

402 = 2 + (n-1)*4

We'll remove the brackets:

402 = 2 + 4n - 4

We'll combine like terms:

402 = 4n - 2

We'll add 2 both sides:

4n = 404

We'll divide by 4:

n = 101

So, the number of terms, from 2 to 402 is n = 101 terms.

S101 = (2 + 402)*101/2

S101 = 404*101/2

S101 = 202*101

**So the sum of the terms of the arithmetic progression is **

**S101 = 20402**