# nth Term of an AP

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Consider the following AP:

$2,\,\,5,\,\,8,\,\,11,\,\,13...$

The first term a of this AP is 2, the second term is 5, the third term is 8, and so on. We write this as follows:

$\begin{array}{l}{T_1} = a = 2\\{T_2} = 5\\{T_3} = 8\\...\end{array}$

The n-th term of this AP will be denoted by Tn. How can you find the n-th term for any value of n? For example, what will be the value of the following terms?

${T_{20}},\,\,{T_{45}},\,\,{T_{90}},\,\,{T_{200}}$

Obviously, we cannot evaluate each and every term of the AP to determine these specific terms. Instead, we must develop a relation which enables us to find the n-term for any value of n.

To do that, consider the following relations for the terms in an AP:

$\begin{array}{l}{T_1} = a\\{T_2} = a + d\\{T_3} = a + d + d = a + 2d\\{T_4} = a + 2d + d = a + 3d\\{T_5} = a + 3d + d = a + 4d\\{T_6} = a + 4d + d = a + 5d\\...\end{array}$

What pattern do you observe? If you have to calculate the sixth term, for example, then you have to add five times d to the first term a. Similarly, if you have to calculate the n-th term, how many times will you have to add d to a? The answer should be easy: one less than n. Thus,

${T_n} = a + \left( {n - 1} \right)d$

This relation helps us calculate any term of an AP, given its first term and its common difference. Thus, for the AP above, we have:

$\begin{array}{l}{T_{20}} = 2 + \left( {20 - 1} \right)3 = 2 + 57 = 59\\{T_{45}} = 2 + \left( {45 - 1} \right)3 = 2 + 132 = 134\\{T_{90}} = 2 + \left( {90 - 1} \right)3 = 2 + 267 = 269\\{T_{200}} = 2 + \left( {200 - 1} \right)3 = 2 + 597 = 599\end{array}$

## Slide-4: Examples

Example 1: Find the 20th term of an AP whose 3rd term is 5 and 7th term is 13.

Solution: We denote by a and d the first term and the common difference respectively of the AP. We have:

$\begin{array}{l}{T_3} = a + 2d = 5\\{T_7} = a + 6d = 13\\ \Rightarrow \,\,\,4d = 8\\ \Rightarrow \,\,\,d = 2\\ \Rightarrow \,\,\,a + 2\left( 2 \right) = 5\\ \Rightarrow \,\,\,a = 1\end{array}$

Now that we have a and d, we can calculate any term:

${T_{20}} = a + 19d = 1 + 19\left( 2 \right) = 39$

Example 2: How many three-digit numbers are divisible by 3?

Solution: The smallest three-digit number which is divisible by 3 is 102. The largest three-digit number divisible by 3 is 999. Clearly, we have to find the number of terms in the following AP:

$102,\,\,105,\,\,108,\,\,...\,,999$

Let 999 be the n-th term in this AP. We note that a is equal to 102, and d is equal to 3. Thus,

$\begin{array}{l}{T_n} = a + \left( {n - 1} \right)d = 102 + \left( {n - 1} \right)3 = 999\\ \Rightarrow \,\,\,3\left( {n - 1} \right) = 999 - 102 = 897\\ \Rightarrow \,\,\,n - 1 = \frac{{897}}{3} = 299\\ \Rightarrow \,\,\,n = 300\end{array}$

We see that there are 300 three-digit numbers which are divisible by 3.

Example 3: Consider the following AP:

$7,\,\,11,\,\,15,\,\,19,\,...$

Is the number 301 a part of this AP?

Solution: Note that a is equal to 7 and d is equal to 4. Let 301 be the n-term of this AP, where n is a positive integer. We have:

$\begin{array}{l}{T_n} = a + \left( {n - 1} \right)d = 7 + \left( {n - 1} \right)4 = 301\\ \Rightarrow \,\,\,4\left( {n - 1} \right) = 301 - 7 = 294\\ \Rightarrow \,\,\,n - 1 = \frac{{294}}{4} = \frac{{147}}{2}\\ \Rightarrow \,\,\,n = \frac{{149}}{2}\end{array}$

We have obtained n as a non-integer, whereas n should have been an integer. This can only mean that 301 is not part of the given AP. In fact, you can verify that the numbers 299 and 303 are part of this AP.

Example 4: Consider the following AP:

$10,\,\,7,\,\,4,\,...\,, - 62$

Find the 13th term from the last term (towards the first term) of this AP.

Solution: To solve this question, we consider the given AP in reverse:

$- 62,\,\, - 59,\,\, - 56,\,...\,,4,\,\,7,\,\,10$

Seen this way, we have:

$a = - 62,\,\,d = 3$

Thus,

$\begin{array}{l}{T_{13}} = a + 12d = - 62 + 12 \times 3\\ \;\;\;\;\;\;\,\, = - 62 + 36\\ \,\,\,\,\,\,\,\,\,\,\,\,= - 26\end{array}$

Example 5: Consider the following two APs:

$\begin{array}{l}3,\,\,13,\,\,23,\,\,33,...\\1,\,\,8,\,\,15,\,\,22,...\end{array}$

What is the first term in the first AP which also occurs in the second AP?

Solution: Let the term we seek be x, and let it be the m-th term in AP-1 and the n-th term in AP-2. For AP-1, a is 3 and d is 10. For AP-2, a is 1 and d is 7. Thus, we have:

$\begin{array}{l}x = 3 + \left( {m - 1} \right)10 = 10m - 7\\x = 1 + \left( {n - 1} \right)7 = 7n - 6\\ \Rightarrow \,\,\,10m - 7 = 7n - 6\\ \Rightarrow \,\,\,m = \frac{{7n + 1}}{{10}}\end{array}$

We need to think of the smallest m which will satisfy this relation, given that both m and n are integers. A little thinking will show that the values we are looking for are:

$m = 5,\,\,\,n = 7$

Thus, the fifth term of AP-1 is the seventh term of AP-2, and this term is 43.

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