**Introduction **

In the 19^{th} century in Germany, a Math class for grade 10 was going on.

The teacher asked her students to sum all the numbers from \(1\) up to \(100\).

The students were struggling to calculate the sum of all these numbers.

One boy shouted out the answer \(5050\) while the other students were still in the initial steps of calculating the sum.

This boy was the great German mathematician Carl Friedrich Gauss.

How did he arrive at the sum so quickly?

Well, he noticed that terms equidistant from the beginning and the end of the series had a constant sum equal to \(101\).

We can see that in the sequence \(1,2,3,...,100\), there are \(50\) such pairs whose sum is \(101\).

Thus, the sum of all terms of this sequence is:

\[ 50 \times 101 = 5050\]

**Sum of n Terms of an Arithmetic Progression **

Consider an arithmetic progression (AP) whose first term is \(a_1\) (or) \(a\) and the common difference is \(d\).

- The sum of the first \(n\) terms of an arithmetic progression when the \(n^{th} \) term is NOT known is:

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

- The sum of the first \(n\) terms of an arithmetic progression when the \(n^{th} \) term, \(a_n\) is known is:

\[ S_n=\frac{n}{2}[a_1+a_n] \]

How do we derive these formulas?

We will use the same logic used above by Carl Friedrich Gauss.

Let us consider the arithmetic progression with \(n\) terms:

\[a,a+d,a+2d,...(a+(n-2) d),(a+(n-1) d)\]

The sum of \(n\) terms of this progression is:

\[ S_n=a+(a+d)+\ldots+(a+(n-2) d)+(a+(n-1) d) \,\,\,\,\,\, \rightarrow (1) \]

By reversing the order of the terms of this equation:

\[ S_n=(a+(n-1) d)+(a+(n-2) d)+\ldots+(a+d)+a \,\,\,\,\,\, \rightarrow (2) \]

We see that the sum of corresponding terms of equation (1) and equation (2) yield the same sum which is \(2a+(n-1)d\).

We know that there are totally \(n\) terms in the above AP.

So by adding (1) and (2), we get:

\[ \begin{align}

2S_n &= n(2a+(n-1) d)\\[0.3cm]

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

\end{align} \]

The above sum of arithmetic progression equation can be written as:

\[ \begin{align}

S_n &= \frac{n}{2}(2a+(n-1)d)\\[0.3cm]

S_n &= \frac{n}{2}(a+a+(n-1)d)\\[0.3cm]

S_n &= \frac{n}{2}(a_1+a_n) [\because a_n=a+(n-1)d \text{ and } a=a_1 ]

\end{align} \]

Thus, the sum of arithmetic progression equations are:

\(\begin{align} S_n &= \frac{n}{2}(2a+(n-1)d)\\[0.3cm] S_n &= \frac{n}{2}(a_1+a_n) \end{align} \) |

Let's take a look at the following flowchart to get an idea of the formula that has to be used to find the sum of arithmetic progression according to the information available to us.

**Sum of AP of Natural Numbers**

The AP of natural numbers is:

\[1,2,3,...,n,...\]

Let us find the sum of \(n\) terms of AP of natural numbers.

It would be:

\[1+2+3+...+n\]

We can find the sum in two methods using the above two formulas.

**Method 1**

Here,

The first term is, \(a=1\).

The common difference is, \(d=1\).

The number of terms is \(n\).

Substitute all these values in the first sum of AP formula:

\[\begin{align}

S_n &= \frac{n}{2}(2a+(n-1)d)\\[0.3cm]

S_n &= \frac{n}{2}(2(1)+(n-1)1) \\[0.3cm]

S_n&= \frac{n}{2}(2+n-1)\\[0.3cm]

S_n&= \frac{n(n+1)}{2}

\end{align} \]

**Method 2**

Here,

The first term is, \(a_1=1\).

The nth term of the above AP is, \(a_n=n\).

Substitute all these values in the second sum of AP formula:

\[\begin{align}

S_n &= \frac{n}{2}(a_1+a_n)\\[0.3cm]

S_n&= \frac{n}{2} (1+n)\\[0.3cm]

S_n&= \frac{n(n+1)}{2}

\end{align} \]

Thus, from the above methods, the sum of AP of natural numbers is:

Sum of AP of natural numbers \( = \dfrac{n(n+1)}{2} \) |

**Sum of Infinite AP **

Let us consider an example for the sum of an infinite AP.

\[ 2+ 5 + 8+... \]

Here, the first term is, \(a=2\).

The common difference is, \(d=3\).

The number of terms is, \(n= \infty\).

Substitute all these values in the formula of sum of AP:

\[\begin{align}

S_n &= \frac{n}{2}(2a+(n-1)d)\\[0.3cm]

S_n &= \frac{\infty}{2}(2(2)+(\infty-1)3) \\[0.3cm]

S_n&= \infty

\end{align} \]

We found the sum of infinite AP to be \(\infty\) when

\(d>0\).

In the same way, the sum of infinite AP is \(-\infty\) when

\(d<0\).

Thus,

\(\text { Sum of infinite } A P=\left\{\begin{array}{ll} \infty, & \text { if } \quad d>0 \\[0.3cm] -\infty, & \text { if } \quad d<0 \end{array}\right. \) |

- The sum of arithmetic progression whose first term is \(a\) and the common difference is \(d\) can be calculated using one of the following formulas:

\[\begin{align}

S_n &= \frac{n}{2}(2a+(n-1)d)\\[0.3cm]

S_n &= \frac{n}{2}(a_1+a_n)

\end{align} \] -
The sum of AP of \(n\) natural numbers is:

\[ \dfrac{n(n+1)}{2} \] -
Sum of infinite AP is:

\[\left\{\begin{array}{ll}

\infty, & \text { if } \quad d>0 \\[0.3cm]

-\infty, & \text { if } \quad d<0

\end{array}\right.

\]

**Sum of AP Calculator **

We can find out the sum of AP using the following “Sum of AP Calculator” by entering the first term, common difference, and the number of terms.

**Solved Examples **

Example 1 |

Calculate the following sum:

\[S=\underbrace{190+167+144+121+\ldots}_{20 \text { terms }} \]

**Solution:**

We do not know the last term in this sequence, so we will use the first formula to calculate this sum, which is:

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

Here, we have:

\[a=190, \quad d=-23, \quad n=20 \]

Substituting all these values in the above formula,

\[ \begin{align}S &=\frac{20}{2}(2(190)+(20-1)(-23)) \\[0.3cm]

&=10(380-437) \\[0.3cm]

&=10(-57) \\[0.3cm]

&=-570

\end{align} \]

\(\therefore S=-570\) |

Example 2 |

Consider the following AP:

\[24,21,18, \ldots \]

How many terms of this AP must be considered so that their sum is \(78\)?

**Solution:**

Let the number of terms that give the sum \(78\) be denoted as \(n\).

We have:

\[a=24, d=-3, \quad S=78 \]

Substituting all these values in the first formula of sum of AP,

\[\begin{aligned}

S=& \frac{n}{2}(2 a+(n-1) d) \\[0.3cm]

\Rightarrow & 78=\frac{n}{2}(48-3(n-1)) \\[0.3cm]

&=\frac{n}{2}(51-3 n) \\[0.3cm]

\Rightarrow & 3 n^{2}-51 n+156=0 \\[0.3cm]

\Rightarrow & n^{2}-17 n+52=0 \\[0.3cm]

\Rightarrow &(n-4)(n-13)=0 \\[0.3cm]

\Rightarrow & n=4, \,\,13

\end{aligned}

\]

\(\therefore \begin{align} \text{The sum of 4 terms} &=78\\ \text{The sum of 13 terms} &=78\\ \end{align} \) |

Example 3 |

Given \(a=5\), \(d=3\) and \(a_n=50\), find the value of \(S_n\).

**Solution:**

The given values are:

\[ \begin{align}

a&=5=a_1\\

d&=3\\

a_n &=50

\end{align} \]

We know that the \(n^{th}\) term of AP is given by the formula:

\[ \begin{align}

a_n&= a+(n-1)d\\[0.3cm]

50 &= 5+ (n-1) 3 \\[0.3cm]

50&= 5+3n-3\\[0.3cm]

50&= 3n+2\\[0.3cm]

48&=3n\\[0.3cm]

16&=n

\end{align} \]

Since we know the \(n^{th}\) term of AP, we use the following formula to find the sum:

\[\begin{align} S_n &= \frac{n}{2}(a_1+a_n) \\[0.3cm] S_n&= \frac{16}{2}(5+50) \\[0.3cm] S_n&= 8 (55) \\[0.3cm] S_n&=440\end{align}\]

\(\therefore S_n = 440\) |

- Find the sum \(7+10 \frac{1}{2}+14+\ldots+84\).
- Find the sum \( \frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots, \text { to } 13 \text { terms. } \)
- How many terms of the AP \(9,17,25, \ldots \) . must be taken to give a sum of \(636\)?

**Practice Questions**

**Maths Olympiad Sample Papers**

IMO (International Maths Olympiad) is a competitive exam in Mathematics conducted annually for school students. It encourages children to develop their math solving skills from a competition perspective.

You can download the FREE grade-wise sample papers from below:

- IMO Sample Paper Class 1
- IMO Sample Paper Class 2
- IMO Sample Paper Class 3
- IMO Sample Paper Class 4
- IMO Sample Paper Class 5
- IMO Sample Paper Class 6
- IMO Sample Paper Class 7
- IMO Sample Paper Class 8
- IMO Sample Paper Class 9
- IMO Sample Paper Class 10

To know more about the Maths Olympiad you can **click here**

**Frequently Asked Questions (FAQs)**

## 1. What is the sum of an arithmetic sequence?

The sum of an arithmetic sequence is “the sum of the first \(n\) terms” of the sequence and it can found using one of the following formulas:

\[\begin{align}

S_n &= \frac{n}{2}(2a+(n-1)d)\\[0.3cm]

S_n &= \frac{n}{2}(a_1+a_n)

\end{align} \]

Here,

\(a=a_1\)=the first term

\(d\) = the common difference

\(n\)= number of terms

\(a_n=n^{th} \) term

\(S_n \) = the sum of the first \(n\) terms.

## 2. What is the sum of n terms of AP?

The sum of an arithmetic sequence is “the sum of the first \(n\) terms” of the sequence and it can found using one of the following formulas:

\[\begin{align}

S_n &= \frac{n}{2}(2a+(n-1)d)\\[0.3cm]

S_n &= \frac{n}{2}(a_1+a_n)

\end{align} \]

Here,

\(a=a_1\) = the first term

\(d\) = the common difference

\(n\) = number of terms

\(a_n=n^{th} \) term

\(S_n \) = the sum of the first \(n\) terms.

Among these two formulas, the first formula is the most frequently used sum of n terms of AP formula.

## 3. What is the sum of a series?

We use multiple formulas while finding the sum of a series.

- If the series is an AP, then we use the following formulas to find the sum.

\[\begin{align}

S_n &= \frac{n}{2}(2a+(n-1)d)\\[0.3cm]

S_n &= \frac{n}{2}(a_1+a_n)

\end{align} \]

Here,

\(a=a_1\) = the first term

\(d\) = the common difference

\(n\) = number of terms

\(a_n=n^{th} \) term

\(S_n \) = the sum of the first \(n\) terms.

- If the series is GP, then we use the following formula to find the sum.

\[ S_{n}=\frac{a\left(r^{n}-1\right)}{r-1} \]

Here,

\(a\) = the first term

\(r\) = the common ratio

\(n\) = number of terms

\(S_n \) = the sum of the first \(n\) terms.

## 4. What is the formula of \(S_n\)?

We use multiple formulas while finding the sum of a series that is represented by \(S_n\).

- If the series is AP, then we use the following formulas to find the sum.

\[\begin{align}

S_n &= \frac{n}{2}(2a+(n-1)d)\\[0.3cm]

S_n &= \frac{n}{2}(a_1+a_n)

\end{align} \]

Here,

\(a=a_1\) = the first term

\(d\) = the common difference

\(n\)= number of terms

\(a_n=n^{th} \) term

\(S_n \) = the sum of the first \(n\) terms.

- If the series is GP, then we use the following formula to find the sum.

\[ S_{n}=\frac{a\left(r^{n}-1\right)}{r-1} \]

Here,

\(a\) = the first term

\(r\) = the common ratio

\(n\) = number of terms

\(S_n \) = the sum of the first \(n\) terms.