nth Term of an AP: Formula, Definition and Examples

nth term of an AP: According to a definite rule, some numbers are arranged in a definite order to form a sequence. The number occurring at the \({n^{th}}\) place is called the nth term of an AP, denoted by \({T_n}\) or \({a_n}\). A sequence in which each term differs from its preceding term by a constant is called an arithmetic progression, written as AP. 

This constant is called the common difference of the AP. If the first term of an arithmetic progression is \(a\) and its common difference is \(d\), then the \({n^{th}}\) term can be found. So, in this article, we will understand what is nth Term of AP and also learn more through the examples of the \({n^{th}}\) term of an arithmetic progression (AP).

What is a Sequence?

An arrangement of numbers in which one number is designated as first, another as the second, another as the third, and so on is known as a sequence.

Consider the following arrangement of numbers:

\(1, 4, 9, 16, 25, …….. \)
\(1, 3, 5, 7, 9, ……\)

In the above arrangements, numbers are arranged in a definite order according to some rules. The numbers are squares of natural numbers in the first arrangement, and in the second arrangement, the numbers are odd natural numbers. Each of the above formats is a sequence. 

So, a sequence is an arrangement of numbers in a definite order according to some rule.

The various numbers occurring in a sequence are called its terms. We denote the terms of a sequence by \({x_1},{x_2},{x_3}, \ldots \ldots \ldots ,{x_n}\), etc. The subscripts denote the position of the term. The first number or the number at first place is called its first term of the sequence and indicated by \({x_1}\), the second number is called the second term and denoted by \({x_2}\) and so on.

The \({n^{th}}\) term is the number at the \({n^{th}}\)position of the sequence and is denoted by \({x_n}\). The \({n^{th}}\) term is also called a general term of the sequence.

For example, \(3, 6, 9, 12, …….\) is a sequence whose first term is \(3\), i.e., \({x_1} = 3\), the second term is \(6\), i.e., \({x_2} = 6\), the third term is \(9\), i.e., \({x_3} = 9\) the fourth term is \(12\), i.e., \({x_4} = 12\) and so on.

What is a Progression?

If the terms of a sequence are written under fixed conditions, then the sequence is called a progression. There are three types of progression. They are arithmetic progression, geometric progression, and harmonic progression.

Here we will study a particular type of progression called Arithmetic Progression (AP) or Arithmetic Sequence and the \({n^{th}}\) term of an AP.

Arithmetic Progression

Consider the following sequences.

1. \(1, 3, 5, 7, 9, ……..\)
2. \(4, 8, 16, 24, …….\)
3. \(10, 20, 30, 40, ……\)

In the above sequences, every term except the first is obtained by adding a fixed number (positive or negative) to the preceding term. For example, in the sequence given in (i), each term is obtained by adding \(2\) to the prior term. In the sequence given in (ii), each term is \(4\) more than the preceding term, and in the sequence provided in (iii), each term is obtained by adding \(10\) to the prior term.

All these sequences are called arithmetic sequences or arithmetic progressions abbreviated as AP. 

A sequence \({a_1},{a_2},{a_3}, \ldots \ldots \ldots ,{a_n}\) is called an arithmetic progression, if there exists a constant term d such that

\({a_2} = {a_1} + d\)
\({a_3} = {a_2} + d\)
\({a_4} = {a_3} + d\)
\({a_n} = {a_{n – 1}} + d\) and so on.

There are two types of AP they are, finite and infinite AP.

Finite AP: An AP having a finite number of terms is called a finite AP. A finite AP has the last term. 

Example: \(2, 6, 10, 14, 18, 22, 26, 30\).

Here, the first term, \({a_1} = 2\)
The common difference, \(d=6-2=10-6=14-10=4\)
Number of terms, \(n=8\)
Last term, \({a_n} = {a_8} = 30\)

Infinite AP: An infinite is the AP which is not having a fixed number of terms. In other words, an AP which has an infinite number of terms is called an infinite AP.

Example: \(1, 2, 3, 4, 5, 6, 7, ………\)

Here, the first term, \({a_1} = 1\)
The common difference, \(d=2-1=3-2=4-3=1\)
Number of terms, \(n=∞\), or we cannot define the number of terms
Last term, \({a_n}\) also cannot be found out.

nth term of an AP Formula

In this section, we will find what is nth Term of AP using the formula for the \({n^{th}}\) term or general term of an AP in terms of its first term and the common difference. The same will be used to solve some problems on AP.

Let \(a\) be the first term, and \(d\) be the common difference of an AP. Then, its \({n^{th}}\) term or general term is given by

\({a_n} = a + (n – 1)d\)

Let us see how this is obtained.

Let \({a_1},{a_2},{a_3}, \ldots \ldots .{a_{n \cdot }}\) be the given AP. Then, 

\({a_1} = a\)
\( \Rightarrow {a_1} = a + (1 – 1)d \ldots \ldots \)(i)

Since each term of an AP is obtained by adding a common difference to the preceding term.
Therefore, \({a_2} = a + d\)
\( \Rightarrow {a_2} = a + (2 – 1)d\)(ii)

Similarly, we have \({a_3} = {a_2} + d\)
\( \Rightarrow {a_3} = (a + d) + d\)
\( \Rightarrow {a_3} = a + 2d\)
\( \Rightarrow {a_3} = a + (3 – 1)d\)……(iii)

and \({a_4} = {a_3} + d\)

\( \Rightarrow {a_4} = (a + 2d) + d\)
\( \Rightarrow {a_3} = (a + 3d)\)
\( \Rightarrow {a_3} = a + (4 – 1)d\) …….(iv)

Observing the pattern in equation (i), (ii), (iii), and (iv), we find that 

\({a_n} = a + (n – 1)d\)

Therefore, the general term of an AP \(=\) First term \(+\) (term number\(-1\)) \(\times\) common difference

nth Term of an AP from the End

Let there be an AP with first term \(a\) and common difference \(d\). If there are \(m\) terms in the AP, then 

\({n^{th}}\) term from the end \( = {(m – n + 1)^{th}}\) term the beginning
\( \Rightarrow {n^{th}}\) term from the end \( = {a_{m – n + 1}}\)
\(\Rightarrow {n^{th}}\) term from the end \( = a + (m – n + 1 – 1)d\)
\( \Rightarrow {n^{th}}\) term from the end \( = a + (m – n)d\)

Also, if \(l\) is the last term of the AP, then \({n^{th}}\) term from the end is the \({n^{th}}\) term of an AP whose first term is l and the common difference is \(-d\).

Therefore,\({n^{th}}\) term from the end = last term \( + (n – 1)( – d)\)

\(\Rightarrow {n^{th}}\) term from the end \(= l – (n – 1)d\)

Selection of Terms in an AP

The following ways of selecting terms are generally very convenient to solve the problems in AP.

i). Suppose there are three terms in an AP. Then the general terms that must be selected are \(a-d, a, a+d\) with the common difference \(d\).

ii). If there are four terms in an AP, then the general terms that must be selected are \(a-3d, a-d, a+d, a+3d\)

iii). If there are five terms in an AP, then the general terms that must be selected are \(a-2d, a-d, a, a+d, a+2d\).

iv). If there are six terms in an AP, then the general terms selected are \(a-5d, a-3d, a-d, a+d, a+3d, a+5d\) and the common difference is \(2d\).

v). In the case of an odd number of terms, the middle term is a, and the common difference is \(d\) while in the case of an even number of terms, the middle terms are \(a — d, a + d\), and the common difference is \(2d\).

Solved Examples – nth Term of an AP

Q.1: If \(m\) times the \({m^{th}}\) term of an AP is equal to n times its \({n^{th}}\) term, show that the \({(m + n)^{th}}\) term of the AP is zero.
Ans: Let \(a\) be the first term and \(d\) be the common difference of the given AP. Then, 
m times the \({m^{th}}\) term \(=\) n times its \({n^{th}}\) term
\( \Rightarrow m{a_m} = n{a_n}\)
\( \Rightarrow m\left\{ {a + \left( {m – 1} \right)d} \right\} = n\left\{ {a + \left( {n – 1} \right)d} \right\}\)
\( \Rightarrow m\left\{ {a + \left( {m – 1} \right)d} \right\} – n\left\{ {a + \left( {n – 1} \right)d} \right\} = 0\)
\( \Rightarrow a\left( {m – n} \right) + \left\{ {m\left( {m – 1} \right) – n\left( {n – 1} \right)} \right\}d = 0\)
\( \Rightarrow a\left( {m – n} \right) + \left\{ {\left( {{m^2} – {n^2}} \right) – \left( {m – n} \right)} \right\}d = 0\)
\( \Rightarrow a\left( {m – n} \right) + \left\{ {\left( {m – n} \right)\left( {m + n – 1} \right)} \right\}d = 0\)
\( \Rightarrow \left( {m – n} \right)\left\{ {a + \left( {m + n – 1} \right)d} \right\} = 0\)
\( \Rightarrow a + \left( {m + n – 1} \right)d = 0\)
\( \Rightarrow {a_{m + n}} = 0\)
Hence, \({(m + n)^{th}}\)term of the AP is zero.

Q.2: Determine the general term of an AP whose \({7^{th}}\) term is \(-1\) and \({16^{th}}\) term is \(17\).
Ans: Let \(a\) be the first term and \(d\) be the common difference of the given AP.
We know that, \({a_n} = a + (n – 1)d\)
So, \({a_7} = a + (7 – 1)d\) and \({a_{16}} = a + (16 – 1)d\)
Given that \(a + (7 – 1)d = – 1\) and \(a + (16 – 1)d = 17\)
\( \Rightarrow a + 6d = – 1\)…..(i) and
\(a + 15d = 17.\)……(ii)
Subtracting equation (i) from (ii), we get
\(9d = 18 \Rightarrow d = 2\)
Substituting \(d=2\) in equation (i), we get
\(a + 12 = – 1 \Rightarrow a = – 13\)
Hence, the general term \({a_n} = a + (n – 1)d = – 13 + (n – 1)2 = 2n – 15\)

Q.3: If the \({10^{th}}\) term of an AP is \(52\) and \({17^{th}}\) term is \(20\) more than the \({13^{th}}\) term, find the AP.
Ans: Let \(a\) be the first term and \(d\) be the common difference of the given AP.
We have \({a_{10}} = 52\) and \({a_{17}} = {a_{13}} + 20\)
\(\Rightarrow a + 9d = 52\) and \(a + 16d = a + 12d + 20\)
\(\Rightarrow a + 9d = 52\) and \(4d = 20\)
\(\Rightarrow a + 9d = 52\) and \(d = 5\)
\( \Rightarrow a + 9(5) = 52\)
\(\Rightarrow a + 45 = 52\)
\( \Rightarrow a = 7\)
Hence the AP is \(7,12,17,22, \ldots \ldots \)

Q.4: Which term of an AP \(5, 15, 25, …..\) will be \(130\) more than its \({31^{st}}\) term?
Ans: Given \(AP =5, 15, 25, ….. \)
Here, \(a=5\) and \(d=10\)
Therefore, \({a_{31}} = a + 30d = 5 + (30 \times 10) = 5 + 300 = 305\)
So, required term \(=305+130=435\)
Let this be the \({n^{th}}\) term.
Then, \({a_n} = 435\)
\( \Rightarrow 5 + (n – 1) \times 10 = 435\)
\( \Rightarrow 10n = 440\)
\( \Rightarrow n = 44\)
Hence, the \({44^{{\rm{th }}}}\) term will be \(130\) more than its \({31^{st}}\) term.

Q.5: How many terms are there in the sequence \(3, 6, 9, ……111\)?
Ans: Here, \(a=3\) and \(d=3\). Let there be \(n\) terms in the given sequence. Then,
\({n^{th}}\) term \(=111\)
\( \Rightarrow a + (n – 1)d = 111\)
\(\Rightarrow 3 + (n – 1) \times 3 = 111\)
\( \Rightarrow n = 37\)
Hence, there are \(37\) terms in the given sequence.

Summary

In this article we learned that a sequence in which each term differs from its preceding term by a constant is called an arithmetic progression. We also learnt the meaning of a sequence, \({n^{th}}\) term of an arithmetic progression and solved some problems based on \({n^{th}}\) term of an AP.

Frequently Asked Questions (FAQs) – nth Term of an AP

The commonly asked questions about nth Term of an AP are answered here:

Students can make use of NCERT Solutions for Maths provided by Embibe for your exam preparation.

We hope this detailed article on the nth term of an AP helps you. Practice more questions and master this concept. If you have any queries regarding this article, reach out to us through the comment section below and we will get back to you as soon as possible.