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a 3 – a 2 = 0 – 3 = – 3

a 4 – a 3 = –3 – 0 = –3

Similarly this is also an AP whose first term is 6 and the common difference is –3.
In general, for an AP a 1 , a 2 , . . ., a n , we have

d = a k+1 – a k
where a k+1 and a k are the (k+1)th and the kth terms respectively.

Example –
Does the following list of numbers does form an AP? If yes, write the next two terms :

4, 10, 16, 22, . . .
Solution- We have, a 2 – a 1 = 10 – 4 = 6

a 3 – a 2 = 16 – 10 = 6

a 4 – a 3 = 22 – 16 = 6

i.e., a k + 1 – a k is the same every time.

So, the given list of numbers forms an AP with the common difference d = 6.

The next two terms are: 22 + 6 = 28 and 28 + 6 = 34.

nth Term of and A.P. –

The nth term of an A.P is given by T n=a + (n−1)d, where a is the first term, d is a
common difference and n is the number of terms.

Let a 1 , a 2 , a 3 , . . . 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 2 + d = (a + d) + d = a + 2d = a + (3 – 1) d

the fourth term a 4 = a 3 + d = (a + 2d) + d = a + 3d = a + (4 – 1) d……

Looking at the pattern, we can say that the nth term a n = a + (n – 1) d.

So, the nth term an of the AP with first term a and common difference d is given by

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