In my previous posts, I discussed about finding nth term of arithmetic progression. In this post, I will be discussing about how to find sum of n terms of arithmetic progression.

Lets take an example. We have an arithmetic progression of the form **1, 4, 7, 10, 13....**

We can easily find nth term of such a sequence by applying formula , where is the nth term, a is the first term of the sequence and d is the common difference of sequence.

**But, now we want to find sum of first n terms of the given sequence. How are we going to do it?**

The formula is: where a is the first term, d is the common difference and is the sum of first n terms of arithmetic progression.

Lets suppose, we want to find sum of first 10 terms of given AP **1, 4, 7, 10, 13.....**

First term = a = 1

Common difference = d = 4 - 1 =3

n = 10

Applying formula, , we get

In this way, we can find sum of n terms of given AP. Now, the question may arise from where did this formula come. You do not need to know derivation of this formula to solve different problems but some of the students may be interested in knowing about more details. So, lets discuss about derivation of the formula that we just used.

We know that any arithmetic progression can be written in a general form:

a + (a+d) + (a+2d) + (a+3d) .... (a+ (n-1)d)

Let their sum be equal to .

Therefore, ** (1)**

We can also write **(1) **in the form

**(2)**

Adding **(1)** and **(2)**, we can say that

.... n times

**(3)**

which is the required formula.

We can also write **(3) **like:

**(4)**

But, we know that [a + (n-1)d] is equal to the nth term of arithmetic progression. Because, we want to find sum of n terms of AP. The nth term here would be considered as the last term of AP. Suppose, the last term is equal to then we can write **(4)** as:

We can use this formula directly to find sum of n terms of AP if we know first term and last term of AP. This formula will be more clearer to you when I will be solving some problems regarding this formula in the coming posts.