What is the sum of the terms of the series 2 6 10 402 S

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

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