Sum of n Terms of an Arithmetic Progression: Definition, Formula, Examples

Sum of n terms of an Arithmetic Progression: When we say that a collection of objects is listed in a sequence, we usually mean that the collection is ordered to have an identified first member, second member, third member, and so on. For example, the amount of money deposited in a bank over several years form a sequence. Sequences following specific patterns are called progressions.

An arithmetic progression or AP, in short, is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term. So, the sum of \(n\) terms of an AP is the sum of the arithmetic progression having \(n\) number of terms in a series.

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What is Sequence?

The list of objects or items arranged sequentially is known as a sequence. Consider the successive quotients that we obtain in the division of \(10\) by \(3\) at different division steps. In this process, in each step of division, we get the quotients as \(3, 3.3, 3.33, 3.333,…\) and so on. These quotients form a sequence.

The various numbers occurring in a sequence are called its terms. We denote the terms of a sequence by \(a_1, a_2, a_3, …., a_n\), etc. The subscripts denote the position of the term. The \(n^{th}\) term is the number at the \(n^{th}\) position of the sequence and is denoted by \(a_n\). The \(n^{th}\) term is also called a general term of the sequence.

What is Series?

Let \(a_1, a_2, a_3, …., a_n\) be a given sequence. Thus the expression \(a_1 + a_2 + a_3 + ….., + a_n\) is called the series associated with the given sequence. The series is finite or infinite, according to as the given sequence is finite or infinite.

Arithmetic Progression

Consider the following sequences:
(i) \(2, 4, 6, 8, ……..\)
(ii) \(3, 6, 9, 12, ……..\)
(iii) \(100, 70, 40, 10, ……..\)
Each of the numbers in the list is called a term.
In case (i), each term is \(2\) more than the term preceding it.
In case (ii), each term is \(3\) more than the term preceding it.
In case (iii), each term is \(30\) less than the term preceding it.

We see that successive terms are obtained by adding or subtracting a fixed number to the preceding terms in all the lists above. Hence, such a list of numbers is said to form an Arithmetic Progression (AP). So, an arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.

This fixed number is called the common difference of the AP. Remember that it can be positive, negative or zero. Let us denote the first term of an AP by \(a_1\), the second term by \(a_2\), the third term by \(a_3, ….., n^{th}\) term by \(a_n\) and the common difference by \(d\)). Then AP becomes \(a_1, a_2, a_3, ……, a_n\).
So, \(a_2 – a_1 = a_3 – a_2 = ……. = a_n – a_{n-1} = d\)

Using the above cases, we can see \(a, a + d, a + 2d, a + 3d,……\) represents an arithmetic progression where \(a\) is the first term and \(d\) the common difference. This is called the general form of an AP.

\(n^{th}\) Term of an AP

Let us consider the example: A sum of \(₹ 1000\) is invested at \(8\%\) simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of \(30\) years making use of this fact.

To answer this, let us first see what is the simple interest for the first two years would be. We know that simple interest \( = \frac{ {P \times R \times T}}{ {100}}\)
So, the interest at the end \(1^{st}\) year \( = \frac{ {1000 \times 8 \times 1}}{ {100}} = ₹ 80\)
The interest at the end of \(2^{nd}\) year \( = \frac{ {1000 \times 8 \times 2}}{ {100}} = ₹ 160\)
The interest at the end of \(3^{rd}\) year \( = \frac{ {1000 \times 8 \times 3}}{ {100}} = ₹ 240\)
Now, looking at the pattern formed above, we can find the interest deposited yearly for the \(15^{th}\) year.

Interest at the end of \(15^{th}\) year \(=\) Interest for the \(14^{th}\) year \(+ ₹ 80\)
\(= ₹[80+(14-1)(80)]+ ₹ 80\)
\(= ₹[80+(13)(80)]+ ₹ 80\)
\(= ₹[(80)(14)]+ ₹ 80\)
\(= ₹[80+(15-1)80] = ₹ 1200\)
\( \Rightarrow {\text{Simple interest of first-year + }}\left({15 – 1} \right) \times {\text{increment of simple interest each year}}\)

In the same way, the simple interest at the end of \(30^{th}\) year would be
\({\text{Simple interest of first-year + }}\left({30 – 1} \right) \times {\text{increment of simple interest each year}}\)
\(= ₹ 80 + (29) ₹ 80\)
\(= ₹ 2400\)
The above example would have given you some idea about how to write the \(15^{th}\) term, or the \(25^{th}\) term, and more generally, the \(n^{th}\) term of the AP.

Let \(a_1, a_2, a_3, ….., a_n\) be an AP whose first term \(a_1\) is a and the common difference is \(d\).
Then,
The second term \(a_2 = a + d = a + (2-1) d\)
The third term \(a_3 = a + 2d = a + (3-1) d\)
………..
Looking at the pattern, we can say that the \(n^{th}\) term
\(a_n = a + (n-1) d\)
\(a_n\) is also called the general term of the AP. If there are \(m\) terms in the AP, then \(a_m\) represents the last term which is sometimes also denoted by \(l\).

Sum of \(n\) Terms of an Arithmetic Progression

Let us consider an example where Ram was asked to find the sum of the positive integers from \(1\) to \(50\). He immediately replied that the sum is \(1275\). Can you guess how did he do?
He wrote: \(S = 1 + 2 + 3 +…….+ 49 + 50\)
And then, reversed the numbers to write \(S = 50 + 49 + 48 +……+ 2 + 1\)

Adding these two, he got
\(2S = (50 + 1) +(49 + 2) + (48 + 3) +……+ (3 + 48) + (2 + 49) + (1 + 50)\)
\(= 51 + 51 +…….+ 51 + 51\)(\(50\) times)
So, \(S = \frac{{50 \times 51}}{2} = 1275\)
Therefore, the sum \(= 1275\)
We will now use the same technique to find the sum of the first n terms of an AP \(a, a + d, a + 2d, a + 3d,……\)

The \(n^{th}\) term of this AP is \(a + (n – 1) d\). Let \(S\) denote the sum of first \(n\) terms of the AP.
So, \(S = a + (a + d) + (a + 2d) +……….+ [a + (n – 1) d]\) …….(i)
Rewriting the sum in reverse order, we get
\(S = [a + (n – 1) d] + [a + (n – 2) d] +……….+ (a + 2d) + (a + d)\) ……..(ii)

Adding (i) and (ii), we get
\(2S = [2a + (n – 1)d] + [2a + (n – 1)d] + [2a + (n – 1)d] +…….\)(\(n\) times)
or, \(2S = n[2a + (n – 1)d]\) (Since there are \(n\) terms)
or, \(S = \frac{n}{2} [2a + (n – 1)d]\)
So, the sum of the first \(n\) terms of an AP is given by
\(S = \frac{n}{2}\left[{2a + \left({n – 1} \right)d} \right]\)
We can also write it as
\(S = \frac{n}{2}\left[{a + {a_n}} \right]\)
or
\(S = \frac{n}{2}\left[{a + l} \right]\)
Where \(a\) is the first term and \(a_n\) or \(l\) is the last term

Solved Examples

Q.1. Find the sum of the first \(22\) terms of the AP: \(8, 3, – 2,….\)
Ans: Given AP: \(8, 3, – 2,….\)
Here \(a = 8\), common difference \(d = 3 – 8 = – 5\) and \(n = 22\)
We know that, \(S = \frac{n}{2} [2a + (n – 1)d]\)
So, \( {S_ {22}} = \frac{ {22}}{2}[2 \times 8 + (22 – 1) \times ( – 5)]\)
\( \Rightarrow {{\text{S}}_{22}} = 11[16 + (21) \times ( – 5)]\)
\( \Rightarrow {{\text{S}}_{22}} = 11[16 – 105]\)
\( \Rightarrow {{\text{S}}_{22}} = 11(- 89)\)
\( \Rightarrow {{\text{S}}_{22}} = -979\)
So, the sum of the first \(22\) terms of the given AP is \(- 979\).

Q.2. If the sum of first \(14\) terms of an AP is \(1050\) and its first term is \(10\), find the \(20^{th}\) term.
Ans: Given: \(S_{14} = 1050, n = 14\) and \(a = 10\)
We know that, \({S_n} = \frac{n}{2}\left[{2a + \left({n – 1} \right)d} \right]\)
\( \Rightarrow {S_{14}} = \frac{{14}}{2}[(2 \times 10) + (14 – 1)d]\)
\( \Rightarrow 1050 = 7[20 + 13d]\)
\( \Rightarrow 1050 = 140 + 91d\)
\( \Rightarrow 910 = 91d\)
\( \Rightarrow d = 10\)
Therefore, \(a_{20} = 10 + (20-1) 10\)
\( \Rightarrow a_{20} = 200\)
Therefore, \(20^{th}\) term is \(200\).

Q.3. The income of a person is \(₹ 3,00,000\) in the first year, and he receives an increase of \(₹ 10,000\) to his income per year for the next \(19\) years. Please find the total amount he received in \(20\) years.
Ans: Here, \(a = 300000\), \(d = 10000\) and \(n = 20\)
We know that, \(S_n = \frac{n}{2} [2a + (n – 1)d]\)
\( \Rightarrow S_{20} = \frac{20}{2} [2(300000) + (20 – 1)10000]\)
\( \Rightarrow S_{20} =10 [600000 + 190000]\)
\( \Rightarrow S_{20} = 10(790000)\)
\( \Rightarrow S_{20} = 7900000\)
Hence, the person received \(₹ 79,00,000\) as the total amount at the end of \(20\) years.

Q.4. The sum of n terms of two arithmetic progressions are in the ratio \((3n + 8)∶(7n + 15)\). Find the ratio of their \(12^{th}\) terms.
Ans: Let \(a_1, a_2\) and \(d_1, d_2\) be the first terms and common difference of the first and second arithmetic progression, respectively.

According to the given condition, we have
\(\frac{{{\text{Sum to}}\,n\,{\text{terms of the first AP}}}}{{{\text{Sum to}}\,n\,{\text{terms of the second AP}}}} = \frac{{3n + 8}}{{7n + 15}}\)
\( \Rightarrow \frac{{\frac{n}{2}\left[ {2{a_1} + \left( {n – 1} \right){d_1}} \right]}}{{\frac{n}{2}\left[ {2{a_2} + \left( {n – 1} \right){d_2}} \right]}} = \frac{{3n + 8}}{{7n + 15}}\)
\( \Rightarrow \frac{{\left[ {2{a_1} + \left( {n – 1} \right){d_1}} \right]}}{{\left[ {2{a_2} + \left( {n – 1} \right){d_2}} \right]}} = \frac{{3n + 8}}{{7n + 15}}\) ……..(i)
We know that, \(\frac{{{{12}^{{\text{th}}}}{\text{term of first AP}}}}{{{{12}^{{\text{th}}}}{\text{term of second AP}}}} = \frac{{{a_1} + 11{d_1}}}{{{a_2} + 11{d_2}}}\) …….(ii)

Multiply the numerator and denominator by \(2\), we get
\(\frac{{\left[{2{a_1} + 22{d_1}} \right]}}{{\left[{2{a_2} + 22{d_2}} \right]}}\) …….(iii)
On comparing coefficients of common differences of \( \Rightarrow \frac{{\left[ {2{a_1} + \left( {n – 1} \right){d_1}} \right]}}{{\left[ {2{a_2} + \left( {n – 1} \right){d_2}} \right]}}\) and \(\frac{{\left[{2{a_1} + 22{d_1}} \right]}}{{\left[{2{a_2} + 22{d_2}} \right]}}\)
we get \(n – 1 = 22\)
\( \Rightarrow n = 23\)

Substituting \(n = 23\) in (i), we get
\(\frac{{\left[ {2{a_1} + \left( {23 – 1} \right){d_1}} \right]}}{{\left[ {2{a_2} + \left( {23 – 1} \right){d_2}} \right]}} = \frac{{\left( {3 \times 23} \right) + 8}}{{\left( {7 \times 23} \right) + 15}}\)
\(\Rightarrow \frac{{\left[ {2{a_1} + 22{d_1}} \right]}}{{\left[ {2{a_2} + 22{d_2}} \right]}} = \frac{{69 + 8}}{{161 + 15}}\)
\( \Rightarrow \frac{{{a_1} + 11{d_1}}}{{{a_2} + 11{d_2}}} = \frac{{77}}{{176}}\)
\( \Rightarrow \frac{{{a_1} + 11{d_1}}}{{{a_2} + 11{d_2}}} = \frac{{7}}{{16}}\)

Therefore, \(\frac{{{{12}^{{\text{th}}}}{\text{term of first AP}}}}{{{{12}^{{\text{th}}}}{\text{term of second AP}}}} = \frac{7}{{16}}\)

Q.5. Find the sum of the first \(15\) multiples of \(8\).
Ans: The first multiple of \(8\) is \(8\), i.e. \(a = 8\), \(d = 8\) and \(n = 15\)
We know that, \({S_n} = \frac{n}{2}\left[{2a + \left({n – 1} \right)d} \right]\)
\(\Rightarrow {S_{15}} = \frac{15}{2}\left[{2 \times 8 + \left({15 – 1} \right)8} \right]\)
\(\Rightarrow {S_{15}} = \frac{15}{2}\left[{16 + \left({14} \right)8} \right]\)
\(\Rightarrow {S_{15}} = \frac{15 \times 128}{2}\)
\(\Rightarrow {S_{15}} = 960\)
Therefore, the sum of the first \(15\) multiples of \(8\) is \(60\).

Summary

In the above article, we have learned the definition of sequence, series, arithmetic progression and formula to find the \(n^{th}\) term and sum of \(n\) terms of an arithmetic progression and solved some example problems.

Frequently Asked Questions

We have provided some frequently asked questions here:

Q.1. How to find the sum of arithmetic progression?
Ans: The sum of an arithmetic progression of n terms can be found using the below formula.
\({S_n} = \frac{n}{2}\left[{2a + \left({n – 1} \right)d} \right]\)

Q.2. What is the use of arithmetic progression?
Ans: An arithmetic progression is a sequence with consecutive terms having a common difference between succeeding and the preceding terms. It is used to gather a set of patterns that we observe in our day to day life.
For example, AP used in predicting any sequence, like when someone is waiting for a metro train. Assuming that the train is moving at a constant speed, he/she can expect when the train will come.

Q.3. How to find the common difference in an arithmetic progression?
Ans: Common difference is the difference between the succeeding term and the preceding term.
Example: If \(a_1, a_2, a_3, …..\) are in AP, then, common difference \(d = a_2 – a_1\).

Q.4. What is the arithmetic progression formula?
Ans: The general formula for the \(n^{th}\) term of an arithmetic progression is \(a_n = a + (n-1) d\) and for the sum of \(n\) terms is \({S_n} = \frac{n}{2}\left[{2a + \left({n – 1} \right)d} \right]\)

Q.5. What is arithmetic progression? Give an example.
Ans: An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term. This fixed number is called the common difference of the AP. Remember that it can be positive, negative or zero.
Example: \(1, 3, 5, 7,….\) are in arithmetic progression with the first term as \(1\), common difference \(= 3 – 1 = 2\)

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