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

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:

6 - 2 = 10 - 6 = ...... =
4

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