Nth Term of an AP

In Mathematics, an arithmetic progression (AP) is defined as the list or sequence of numbers, in which each term in the sequence is obtained by adding a fixed number to the preceding term. The fixed number is called the common difference of the arithmetic progression. The fixed number can be positive or negative or zero. Generally, the first term of an AP is represented by a₁, the second term by a₂, … and the nth term by a_n. The common difference of an AP is denoted by “d”. In this article, let us discuss how to find the nth term of an AP with many solved examples.

n^th Term of an AP Formula

Assume that a₁, a₂, a₃,… be an arithmetic progression (AP), in which first term a₁ is equal to “a” and the common difference is taken as “d”, then the second term, third term, etc can be calculated as follows:

Second term, a₂ = a+d

Third term, a₃ = (a+d)+d = a+2d,

Fourth term, a₄ = (a+2d)+d = a+3d, and so on.

Therefore, the nth term of an AP (a_n) with the first term “a” and common difference “d” is given by the formula:

n^th term of an AP, a_n = a+(n-1)d.

(Note: The nth term of an AP (a_n) is sometimes called the general term of an AP, and also the last term in a sequence is sometimes denoted by “l”.)

Also, read: Sum of N terms of AP

n^th Term of an AP Examples

Go through the following examples to understand the procedure for finding the nth term of an AP.

Example 1:

Determine the 10th term of an AP 2, 7, 12, ….

Solution:

Given arithmetic progression (AP) is 2, 7, 12, …

Here, the first term, a = 2.

Common difference, d = 7-2 = 5

n=10.

The formula to find the nth term of an AP, a_n = a+(n-1)d

Now, substitute the values in the formula, we get

a₁₀ = 2 + (10-1)5

a₁₀ = 2 + (9)5

a₁₀ = 2+45

a₁₀ = 47.

Hence, the 10th term of an AP 2, 7, 12, … is 47.

Example 2:

The third term of an AP is 5 and the 7th term of an AP is 9. Find the arithmetic progression (AP).

Solution:

Given that, Third term of AP = 5

Seventh term of AP = 9

(i.e) a₃ = a+(3-1)d = a+2d = 5 …(1)

a₇ = a+(7-1)d = a+6d = 9 …(2)

Now, solve the equations (1) and (2), we get

a=3 and d = 1.

Therefore, the first term is 3 and the common difference is 1.

Therefore, the arithmetic progression (AP) is 3, 4, 5, 6, 7, 8, 9, ….

Example 3:

How many two-digit numbers are divisible by 3?

Solution:

The sequence of two-digit numbers which are divisible by 3 are:

12, 15, 18, 21, …, 99.

To find whether the given sequence is an AP, find the common difference.

The common difference (d) of the above-given sequence is 3, and hence, the given sequence is an Arithmetic progression (AP).

Hence, a = 12, d = 3, a_n = 99.

Now, we have to find the value of “n”.

Now, substitute the values in the formula, a_n = a+(n-1)d, we get

99 = 12+(n-1)3

99 = 12+3n-3

99-12+3 = 3n

3n = 90

n= 90/3

n=30.

Hence, there are 30 two-digit numbers that are divisible by 3.

Practice Problems

Solve the following problems:

1. Find the 30th term of an AP: 10, 7, 4, …..
2. Determine the 31st term of an AP, if its 11th term is 38 and its 16th term is 73.
3. Find the missing terms for the following AP: -4, __, __, ___, ___, 6.

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