#### Question

Prove that no matter what the real numbers *a* and *b *are, the sequence with *the n*th term *a* + *nb* is always an A.P. What is the common difference?

#### Solution

In the given problem, we are given the sequence with the *n*^{th} term (`a_n`) as a + nb where *a* and *b* are real numbers.

We need to show that this sequence is an A.P and then find its common difference (*d*)

Here,

`a_n = a + nb`

Now, to show that it is an A.P, we will find its few terms by substituting n = 1, 2, 3

So,

Substituting* n *= 1*, *we get

`a_1 = a + (1)b`

`a_1 = a + b`

Substituting* n *= 2*, *we get

`a_2 = a+ (2)b`

`a_2 = a + 2b`

Substituting* n *= 3*, *we get

`a_3 = a + (3)b`

`a_3 = a + 3b`

Further, for the given to sequence to be an A.P,

Common difference (*d*) = `a_2 - a_1 = a_3 - a_2`

Here

`a_2 - a_1 = a + 2b - a - b`

= b

Also

a_3- a_2 = a + 3b - a - 2b

= b

Since `a_2 - a_1 = a_3 - a_2`

Hence, the given sequence is an A.P and its common difference is d = b