Sum of AP – Derivation, Formula and Applications

What is an arithmetic progression?

Arithmetic progression is a series of numbers with a constant difference between consecutive terms in an arithmetic sequence. For instance, 3, 5, 7, 9, 11, …… is an arithmetic progression with a common difference of 2 between each succeeding and preceding term. Another example of an AP is the sequence of natural numbers whose common difference is 1. This article shed light on the sum of arithmetic progression.

Carl Friedrich Gauss invented a method to find the sum of n natural numbers. However, the truth behind this claim remains unknown as the origin of this method dates back to the 5th century BC. 

Sum of n terms in an AP

Suppose you have an arithmetic progression (AP) in front of you with the first term as a_1 and common difference as d, then you can easily find the sum of n terms.

Then,

a_{1}=a
a_{2}=a+d
a_{3}=a+2 d
a_{4}=a+3 d

\text { and, } a_{n}=a+(n-1) d

Now, sum of AP is \mathrm{S}_{n}=\mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3}+\ldots \ldots \ldots+\mathrm{a}_{n-1}+\mathrm{a}_{n}

S_{n}=a+(a+d)+(a+2 d)+(a+3 d)+\ldots+\{a+(n-2) d\}+\{a+(n-1) d\} \\ \text { equation } 1

After writing equation 1 in reverse order,

S_{n}=\{a+(n-1) d\}+\{a+(n-2) d\}+\{a+(n-3) d\}+\ldots+(a+3 d)+(a+2 d)+(a+d)+ \\ a-\text { equation } 2

Adding equation 1 and equation 2, we get, 

2 S_{n}=\{2 a+(n-1) d\}+\{2 a+(n-1) d\}+\{2 a+(n-1) d\}+\ldots \ldots+\{a+(n-2) d\}

⇒ 2 S_{n}=n\{2 a+(n-1) d\}

∴ S_{n}=\frac{n}{2}\{2 a+(n-1) d\}    ……. Equation 3

Therefore, if you do not know the last term of an AP series, then you can use 

S_{n}=\frac{n}{2}\{2 a+(n-1) d\} to find the sum of AP.

If you are given the last term l,

then you know that [\mathrm{l}=n^{\text {th }} \text { term }=a+(n-1) d

By substituting the value of l in equation 3, we get

\mathrm{S}_{\mathrm{n}}=\frac{n}{2}\{\mathrm{a}+[\mathrm{a}+(\mathrm{n}-1) \mathrm{d}]\}

\mathrm{S}_{\mathrm{n}}=\frac{n}{2}(\mathrm{a+l})

Therefore, the sum of terms in AP when the last term is ‘l’ is \mathrm{S}_{n}=\frac{n}{2}(\mathrm{a+l})

Sum of infinite AP

Consider this series to find the sum of infinite AP,

3 + 6 + 9 + 12 + ………………

Here, first term, a = 3

The common difference, d = 3

and, number of terms, n = ∞

Substituting these values in the sum of AP formula from equation 3,

S_{n}=\frac{n}{2}\{2 a+(n-1) d\}

⇒ S_{n}=\frac{\infty}{2}\{2(3)+(\infty-1) 3\}

∴ S_{n}=\infty

So, when d > 0, then the sum of infinite AP is ∞.

Similarly, when d < 0, then the sum of infinite AP is -∞.

Examples of AP series

Sum of n natural numbers

Consider the first n natural numbers as 1, 2, 3, 4, 5, 6, 7, ……, n

These natural numbers form an AP sequence with a = 1 and d = 1

According to equation 3,

S_{n}=\frac{n}{2}\{2 a+(n-1) d\}

Substituting the values of a and d in this equation, 

S_{n}=\frac{n}{2}\{2(1)+(n-1) 1\}

⇒ S_{n}=\frac{n}{2}(2+n-1)

⇒ S_{n}=\frac{n(n+1)}{2}

So, the arithmetic progression formula for the sum of n natural numbers is S_{n}=\frac{n(n+1)}{2}

Sum of squares of first n natural numbers

Here, we must find S for the squares of the first n natural numbers

S=1^{2}+2^{2}+3^{2}+4^{2}+\ldots \ldots \ldots \ldots . .+(n-1)^2+n^{2}

We will use the identity below 

a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)

replacing a with n and b with (n – 1) in this equation,

n^{3}-(n-1)^{3}=(n-n+1)\left(n^{2}+n(n-1)+(n-1)^{2}\right)
⇒ n^{3}-(n-1)^{3}=1\left(2 n^{2}-n+n^{2}+1-2 n\right)

⇒ n^{3}-(n-1)^{3}=3 n^{2}-3 n+1  equation 4

Similarly using (\text{n}-1)^{\text{th}} and (\text{n}-2)^{\text{th}} term,

(n-1)^{3}-(n-2)^{3}=3(n-1)^{2}-3(n-1)+1

Now using (\text{n}-2)^{\text{th}} and (\text{n}-3)^{\text{th}} terms,

(n-2)^{3}-(n-3)^{3}=3(n-2)^{2}-3(n-2)+1 equation 6

So on,

2^{3}-1^{3}=3(2)^{2}-3(2)+1

1^{3}-0^{3}=3(1)^{2}-3(1)+1 \ldots \ldots \ldots . . \rightarrow last step

By adding all the steps above, we get n^{3}-0^{3}=3 \Sigma n^{2}-3 \Sigma n+n

n^{3}=3 \Sigma n^{2}-\frac{3 n(n+1)}{2}+n (using the formula for the sum of n natural numbers)

3 \Sigma n^{2}=n^{3}+\frac{3 n(n+1)}{2}-n

Taking n as common,

3 \Sigma n^{2}=n\left[n^{2}+\frac{3(n+1)}{2}-1\right]

3 \Sigma n^{2}=\frac{n}{2}\left(2 n^{2}+3 n+1\right)

Upon factorizing the quadratic equation in the RHS, we get,

\Sigma n^{2}=\frac{n}{6}(2 n+1)(n+1)

\Sigma n^{2}=\frac{n}{6}(2 n+1)(n+1)

Therefore, the sum of squares of first n natural numbers is \frac{n}{6}(2 n+1)(n+1)

Sum of cubes of first n natural numbers

Similar to the above derivations, we can also find the formula for the sum of cubes of first n natural numbers. So, without diving deeper into the complexity of equations, the formula for the sum of cubes of first n natural numbers is \left\{\frac{n(n+1)}{2}\right\}^{2}

Solved examples

Example 1

Find the sum of squares of the first 10 natural numbers.

Solution

The sum of squares of first 10 natural numbers

\text {i e., } 1^{2}+2^{2}+3^{2}+4^{2}+\ldots \ldots \ldots \ldots \ldots \ldots . .+10^{2}

Here, n = 10

And the formula to find the sum is \frac{n}{6}(2 n+1)(n+1)

Substituting the value of n in this formula

We get

=\frac{10}{6}(2(10)+1)(10+1)

=\frac{10}{6}(21)(11)

= 385

Therefore, the sum of squares of the first 10 natural numbers.

Example 2:

Calculate the sum of 10 terms of the series 20, 30, 40, 50, ………..

Solution

Since we do not know the last term of the AP sequence, we will use the formula

S_{n}=a_{2}\{2 a+(n-1) d\}

Here, we have:

a = 20

d = 10

n = 10

by putting these values in the equation, 

S_{10}=\frac{10}{2}\{2(20)+(10-1) 10\}

S_{10}=5\{40+90\}

S_{10}=650

Therefore, the sum of 10 terms of the series 20, 30, 40, 50, ……….. is 650.