Sum of n Terms of an Arithmetic Series: The sum of \(n\) terms in any series is the result of the addition of the first \(n\) terms in that series. In mathematics, series is defined as adding an infinite number of quantities in a specific sequence or order. Hence, a series may also be called an infinite series. In general, a series is represented as:

\(a_{1}+a_{2}+\ldots a_{n}\)

where \(a_{n}\) is the \(n^{\text {th }}\) ordered term.

Let us learn the different types of progression and calculate the sum of \(n\) terms in an arithmetic series.

Sequence vs Series

Sequence and series are important concepts in mathematics. Though they may seem to be the same, they are very different from each other. While a list of numbers written in a specific order is called a sequence, the sum of terms in a sequence is called a series. 

Example: \(1, 2, 3, 4, 5,…\) is a sequence, and \(1 + 2 + 3 + 4 + 5 +…\) is a series. 

Hence, we can say that a sequence is an ordered list of numbers, and series is the sum of numbers in a sequence.

One similarity between sequence and series is that their elements follow a specific progression; they increase or decrease in a logical and predictable pattern.

Types of Series and Sequences

There are different types of progressions followed in sequences and series in mathematics. The four most common types are listed below.

Types of Progressions	Definition
Arithmetic Progression	Every term is formed by adding or subtracting a specific number to the previous term Example: \(2, 4, 6, 8, …\)
Geometric Progression	Every term is formed by multiplying or dividing a specific number to the previous term Example: \(2, 4, 8, 16, …\)
Harmonic Progression	The reciprocals of the terms form an arithmetic sequence Example: \(\frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \ldots\)
Fibonacci Progression	Every term is the sum of the two preceding terms. It begins with \(0\) and \(1\). Fibonacci series: \(0, 1, 1, 2, 3, 5, …\)

In this article, let us understand more about the sum of arithmetic series. 

Mathematician Behind the Sum of n Terms Formula

Gauss was an elementary student in the late \(1700s\). His name was Carl Friedrich Gauss. When his mathematics teacher challenged the students to find the sum of the first \(100\) natural numbers, Gauss amazed everyone by answering correctly in a couple of seconds. The teacher was amazed by his accurate answer as \(5,050\) and further asked him how he calculated the sum. 

Gauss went on to explain that the first \(100\) natural numbers can be divided into fifty pairs such that the first pair is the first number and the last number. The second pair is the second number and the last but one number in the series, and so on.

By this method the pairs of numbers will be \((1+100),(2+99),(3+98), \ldots(50+51)\).

Observe that there are \(50\) pairs, and each pair has a sum of \(101\).

Hence,

\(1+2+3+4+\ldots+100=(1+100),(2+99),(3+98), \ldots(50+51)\)

Therefore, Sum of first \(100\) natural numbers \(=50 \times 51=5,050\)

This method used by Gauss solved the problem instantly, and also led to the generalised formula to find the sum of consecutive numbers.

Observe that the sum of all the pairs is \(101\). While any pair can be used, for ease of use, we consider the first pair. This pair has the first and the last term in the series. Gauss then multiplied the sum by the number of pairs. This number \(50\) is half the total number of terms \(100\). So, to calculate the sum of \(n\) terms in a series, multiply the sum of the first and the last term by half the number of terms in the sequence.

Sum of n Natural Numbers

Natural numbers \(=1+2+3+\ldots n\)

Observe that the natural numbers are in arithmetic progression. Every term is \(1\) more than the previous term. The sum of the first \(n\) terms of an arithmetic series is represented by \(S_{n}\).

\(S_{n}=\frac{n\left(a_{1}+a_{n}\right)}{2}\)

where,

\(n \rightarrow\) number of terms

\(a_{1} \rightarrow\) first term in the series \(=1\)

\(a_{n} \rightarrow\) last term in the series

\(S_{n}=\frac{n\left(1+a_{n}\right)}{2}\)

Sum of n Terms of an Arithmetic Series

Let’s now find the sum of the arithmetic series when the last number \(\left(a_{n}\right)\) is not known.

Consider this arithmetic sequence of \(n\) terms:

First term \(\rightarrow a\)

Second term \(\rightarrow a+d\)

Third term \(\rightarrow a+2 d\)

Fourth term \(\rightarrow a+3 d\)

\(n^{\text {th }}\) term \(\rightarrow a+(n-1) d\)

The sum of \(n\) terms is written as: \(S_{n}=a+(a+d)+(a+2 d)+\ldots(a+(n-1) d)\)

Applying the same concept as used by Gauss, to find the sum of \(n\) terms:

Step 1: Find the sum of the first and the last terms

\(=a+a+(n-1) d\)

\(=2 a+(n-1) d\)

Step 2: Multiply the sum by half the number of terms

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

The sum of \(n\) terms is written as:

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

Flowchart: Sum of n Terms of an Arithmetic Series

Have a look at the flow chart that summarizes the process to find the sum of \(n\) terms in an arithmetic series using the formulas and data available.

Formulas to Calculate Sum of n Terms of an Arithmetic Series

We can use two formulas to calculate the sum of n terms in an arithmetic series.

i. When the first term \(\left(a_{1}\right)\), last term \(\left(a_{n}\right)\) and the number of terms \((n)\) are known,
use the formula:

\(S_{n}=\frac{n\left(a_{1}+a_{n}\right)}{2}\)

ii. When the first term \((a)\), number of terms \((n)\), and the common difference between two consecutive terms \((d)\) are known, use the formula:

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

Sum of an Infinite Arithmetic Series

Consider the infinite series: \(2 + 4 + 6 + 8 +….\)

Here,

First term, \(a=2\)

Number of terms, \(n=\infty\)

Common difference, \(d=4-2=2\)

Sum of arithmetic series, \(S_{n}=\frac{n}{2}(2 a+(n-1) d)\)

\(=\frac{\infty}{2}(2(2)+(\infty-1) \times 2)\)

\(S_{n}=\infty\)

Similarly, for the negative value of common difference, \(S_{n}=-\infty\)

Thus,

Sum of an infinite arithmetic series

\( = \left\{ {\begin{array}{*{20}{c}} \begin{array}{l} \infty ,\\ – \infty , \end{array}&\begin{array}{l} {\rm{if}}\\ {\rm{if}} \end{array}&\begin{array}{l} d > 0\\ d < 0 \end{array} \end{array}} \right.\)

Solved Examples – Sum of n Terms of an Arithmetic Series

Q.1. Find the sum of the arithmetic series up to \(16\) terms: \(1+8+15+22+\ldots\)
Ans: Sum of arithmetic series, \(S_{n}=\frac{n}{2}(2 a+(n-1) d)\)
Here,
\(n=16\)
\(a=1\)
\(d=8-1=7\)
Therefore, \(S_{n}=\frac{16}{2}(2(1)+(16-1) 7)\)
\(=\frac{16}{2}(2+(15) 7)\)
\(=\frac{16}{2}(2+105)\)
\(=8(105)\)
Sum of the arithmetic series up to \(16\) terms \(=840\)

Q.2. Find the sum: \(\sum_{n=1}^{50} 3 n+2\)
Ans: Sum of arithmetic series, \(S_{n}=\frac{n\left(a_{1}+a_{n}\right)}{2}\)
here,
\(n=50\)
\(a_{1}=3(1)+2=5\)
\(a_{20}=3(50)+2=152\)
Therefore, \(S_{n}=\frac{50(5+152)}{2}\)
\(S_{n}=25 \times 157=3,925\)

Q.3. A man saved \(₹16,500\) in ten years. In each year after the first, he saved \(₹100\) more than he did in the preceding year. How much did he save in the first year?
Ans: Total amount saved, \(S_{n}=₹ 16,500 \rightarrow\) sum
Number of years, \(n=10 \rightarrow\) count
Increase in savings every year, \(d= ₹100\) common difference
Sum of arithmetic series, \(S_{n}=\frac{n}{2}(2 a+(n-1) d)\)
Substituting known values,
\(16500=\frac{10}{2}(2 a+(10-1) 100)\)
\(16500=5(2 a+(9) 100)\)
\(16500=5(2 a+900)\)
\(16500=10 a+4500\)
\(10 a=16500-4500\)
\(10 a=12000\)
\(a=\frac{12000}{10}=1200\)
Therefore, savings at the end of the first year, \(a = ₹1,200\)

Q.4. The sum of \(n\) terms of two arithmetic series are in the ratio \(2 n+3: 6 n+5\), then
what is the ratio of their \(13^{\text {th }}\) terms?
Ans: Sum of arithmetic series, \(S_{n}=\frac{n}{2}(2 a+(n-1) d)\)
Let,
First terms \(\rightarrow a_{1}, a_{2}\)
Common differences \(\rightarrow d_{1}, d_{2}\)
Number of terms \(\rightarrow n\)
Therefore, \(\frac{S_{A}}{S_{B}}=\frac{{n}{2}\left(2 a_{1}+(n-1) d_{1}\right)}{\frac{\Delta}{2}\left(2 a_{2}+(n-1) d_{2}\right)}=\frac{2 n+3}{6 n+5}\)
\(\frac{2 a_{1}+(n-1) d_{1}}{2 a_{2}+(n-1) d_{2}}=\frac{2 n+3}{6 n+5}\)
\(\frac{2 a_{1}+(n-1) d_{1}}{2 a_{2}+(n-1) d_{2}}=\frac{2 n+3}{6 n+5}\)
For \(13^{\text {th }}\) term, \(n=13\), for an equivalent value, substitute \(n=25\),
\(\frac{2 a_{1}+(25-1) d_{1}}{2 a_{2}+(25-1) d_{2}}=\frac{2(25)+3}{6(25)+5}\)
\(\frac{2 a_{1}+(24) d_{1}}{2 a_{2}+(24) d_{2}}=\frac{50+3}{150+5}\)
\(\frac{2\left(a_{1}+12 d_{1}\right)}{2\left(a_{2}+12 d_{2}\right)}=\frac{50+3}{150+5}\)
\(\frac{a_{1}+12 d_{1}}{a_{2}+12 d_{2}}=\frac{53}{155} …….(1)\)
\(nth\) term of arithmetic series, \(a_{n}=a+(n-1) d\)
\(13^{\text {th }}\) term of series \(1, \left(\mathrm{a}_{13}\right)_{1}=a_{1}+12 d_{1}\)
\(13^{\text {th }}\) term of series \(2,\left(\mathrm{a}_{13}\right)_{2}=a_{2}+12 d_{2}\)
Ratio of \(13^{\text {th }}\) terms \(=\frac{\left(a_{13}\right)_{1}}{\left(a{13}\right)_{2}}=\frac{a{1}+12 d_{1}}{a_{2}+12 d_{2}} …….(2)\)
Equating \((1)\) and \((2)\),
\(\frac{{{{\left( {{a_{13}}} \right)}_1}}}{{{{\left( {{a_{13}}} \right)}_2}}} = \frac{{53}}{{155}}\)

Q.5. Consider the following series: \(24+21+18+\ldots.\) How many terms of this series must be included to make a sum of \(78\) ?
Ans: Sum of arithmetic series, \(S_{n}=\frac{n}{2}(2 a+(n-1) d)\)
Here,
\(a=24\)
\(d=21-24=-3\)
\(S_{n}=78\)
Therefore, \(78=\frac{n}{2}(2(24)+(n-1) \times-3)\)
\(78=\frac{n}{2}(48-3 n+3)\)
\(78=\frac{n}{2}(51-3 n)\)
\(78=\frac{51 n}{2}-\frac{3 n^{2}}{2}\)
\(78 \times 2=51 n-3 n^{2}\)
\(-3 n^{2}+51 n-156=0\)
\(3 n^{2}-51 n+156=0\)
\(n^{2}-17 n+52=0\)
\(n(n-13)-4(n-13)=0\)
\((n-13)(n-4)=0\)
Therefore,
\(n=3\), and \(14\)
Sum of first \(3\) terms is \(78\)
Sum of first \(14\) terms is \(78\)

Summary

This article starts by contrasting sequence and series. Then, the types of series and progressions are explained with examples. Lastly, the sum of natural numbers and the sum of arithmetic series are explained for first n terms. Again, this is reiterated using a flowchart that explains the steps involved and the decisions to choose the correct formula to find the sum of first n terms in an arithmetic series. Further, the gained knowledge is put to use in the solved problems section, where example problems are solved to find the sum of first n terms in an arithmetic series.

Frequently Asked Questions (FAQs) – Sum of n Terms of an Arithmetic Series

Q.1. What is the formula for the sum of n terms of an arithmetic sequence?
Ans: There are two formulas to calculate the sum of \(n\) terms:
i. When the first term \(\left(a_{1}\right)\), last term \(\left(a_{n}\right)\) and the number of terms \((n)\) are known,
use the formula:
\(S_{n}=\frac{n\left(a_{1}+a_{n}\right)}{2}\)
ii. When the first term \((a)\), number of terms \((n)\), and the common difference between two consecutive terms \((d)\) are known, use the formula:
\(S_{n}=\frac{n}{2}(2 a+(n-1) d)\)

Q.2. What is the sum of an arithmetic progression?
Ans: The sum of numbers in an arithmetic progression is called an arithmetic series.

Q.3. What is the formula for finding the sum of a series?
Ans:
Step 1: Identify the number of terms \((n)\)
Step 2: Identify the first \(\left(a_{1}\right)\) and last \(\left(a_{n}\right)\) values in the series.
Step 3: Substitute the values in the formula for the sum of series: \(S_{n}=\frac{n\left(a_{1}+a_{n}\right)}{2}\)

Q.4. What is the formula for the sum of odd numbers?
Ans: When the first term \((a)\), number of terms \((n)\), and the common difference between two consecutive terms \((d)\) are known, use the formula:
\(S_{n}=\frac{n}{2}(2 a+(n-1) d)\)
For an odd series, \(1+3+5+\ldots\),
\(a \rightarrow\) first term \(=1\)
\(d \rightarrow\) common difference \(=3-1=2\)
\(n \rightarrow\) number of terms
\(\therefore S_{\text {odd }}=\frac{n}{2}(2(1)+(n-1) \times 2)\)
\(=\frac{n}{2}(2+2 n)\)
\(=n \times n\)
Hence, the sum of odd numbers \(=n^{2}\)

Q.5. What is the sum of n natural numbers?
Ans: The sum of \(n\) natural numbers can be calculated using the formula:
\(S_{n}=\frac{n\left(a_{1}+a_{n}\right)}{2}\)
where,
\(n \rightarrow\) number of terms
\(a_{1} \rightarrow\) first term
\(a_{n} \rightarrow\) last term

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