#### Question

The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceed the second term by 6, find three terms.

#### Solution

In the given problem, the sum of three terms of an A.P is 21 and the product of the first and the third term exceeds the second term by 6.

We need to find the three terms

Here,

Let the three terms be (a - d), a, (a + d) where, *a* is the first term and *d* is the common difference of the A.P

So,

`(a - d) + a + (a + d) = 21`

3a = 21

a = 7 .....(1)

Also,

(a - d)(a + d) = a + 6

`a^2 - d^2 = a + 6`

`a^2 - d^2 = a + 6 ("Using " a^2 - b^2 = (a + b)(a - b))`

`(7)^2 - d^2 = 7 + 6` (Using 1)

`49 - 13 = d^2`

Further solving for d

`d^2 = 36`

`d = sqrt36`

d = 6 .....(2)

Now using the values of a and d the expression of the three terms,we get

First term = a - d

So

a - d = 7 - 6

= 1

Second term = a

So

a + d = 7 + 6

= 13

Therefore the three tearm are 1, 7 and 13