General Term of an Arithmetic Progression

Formula for the nth term of the A.P. with first term a and common difference d is a_n = a + (n-1) d.

nth term (a_n) is also called the general term of the AP.

Therefore, General Term of an A.P. = First term + (Term Number – 1) x (Common Difference)

If there are ‘n’ terms in the AP, then a_n represents the last term which is sometimes also denoted by l.

Example1: Find the 14^th, 20^th and nth term of the A.P. given by 9, 13, 17, 21, 25…..

Solution: Here, a (first term) = 9

And,

d = Common difference = 4 (as 13-9 = 4, 17-13=4, 21-17=4 etc.)

We know that a_n = a + (n-1) d

Therefore,

a₁₄ = a + (14-1)d

= a + 13d

= 9 + 13 x 4

= 61

a₂₀ = a + (20-1)d

= a + 19d

= 9 + 19 x 4

= 85

And,

a_n = a+(n-1)d

= 9 + (n-1) x 4

= 4n + 5

Thus, we have

a₁₄ = 61, a₂₀ = 85 and a_n = 4n + 5

Example 2: The first term of an A.P. is -7 and the common difference is 5. Find its 18^th term and the general term.

Solution: We have, a = first term = -7

And

d = common difference = 5

Therefore, a₁₈ = a + (18-1)d

= a + 17d

= -7 + 17 x 5

= 78

And,

a_n = a + (n-1)x5

= -7 + (n-1) x 5

= -7 + 5n -5

= 5n -12

Example 3: Which term of the A.P. 4, 9, 14, 19 … is 299?

Solution: Here, first term (a) = 4 and common difference (d) = 5.

Let consider 299 be the nth term of the given A.P. Then we have,

a_n = 299

a + (n-1)d = 299 (a_n = a + (n-1) d)

4 + (n-1) x 5 = 299

(n-1) x 5 = 299 – 4

n-1 = 295 ÷ 5

n-1 = 59

n = 59 + 1

n= 60

Hence, 60th term of the given sequence is 299.

Example 4: Check whether 125 is a term of the list of numbers 3, 7, 11 ….?

Solution: Clearly, the given sequence is an A.P. with first term a = 3 and common difference d = 4 (7-3=4, 11-7=4).

Let the nth term of the given sequence be 124. Then we have,

a_n = 125

a + (n-1)d = 125

3 + (n-1) x 4 = 125

4n = 125

n = 125/4

Since n is not a natural number. So, 125 is not a term of the given sequence.

Example 5: The 5^th term of an A.P. is 23 and 8^th term is 38. Find the 12^th term and the general term.

Solution: Let us consider (a) be the first term and (d) be the common difference of the given A.P.

Let the A.P. be a₁, a₂, a₃….. a_n

It is given that a₅ = 23 and a₈ = 38

And we know that a_n = a + (n-1) d

So, a + (5-1)d = 23 

a + 4d =23    (i)

And a + (8-1)d = 38

a + 7d = 38   (ii)

Subtracting equation (ii) from equation (i), we get

-3d = -15

d = 5

Putting d =5 in equation (i), we get

a + 4d =23 

a + 4 x 5 = 23

a = 23 - 20

a = 3

Therefore, a₁₂ = a + (12-1)d = 3 + 11x5 = 58 (a = 3, d = 5)

And

a_n = a + (n-1)d = 3 + (n-1)x5 = 5n - 2.

Hence, a₁₂ = 58 and a_n = 5n - 2

Example 6: How many numbers of two digits are divisible by 5?

Solution: We observe that 10 is the first two digit number divisible by 5 and 95 is the last two digit number divisible by 5.

Thus, list of two digit numbers divisible by 5 is 10, 15, 20 ….. 95.

Clearly, it is an A.P. with first term (a) = 10 and common difference (d) = 5 i.e. a = 10 and d = 5

Let us assume there are ‘n’ terms in this A.P. Then,

nth term = 95

a_n = a + (n-1) d

95 = 10 + (n-1) x 5

95 = 10 + 5n – 5

5n = 95-5

n = 90 ÷ 5

n= 18

Hence, there are 18 numbers of two digits which are divisible by 5.

Example 7: Find the11th term from the last term (towards the first term) of the AP: 3, 6, 9 . . ., 111.

Solution: Let’s write the given A.P in the reverse order such that last term becomes the first term.

If we write the given AP in the reverse order, then a = 111 and d = -3

We know that a_n = a + (n-1) d

Therefore,

a₁₁ = 111 + (11-1) x-3

= 111 - 33

= 78

So, the answer is 78.