#### Question

The 4^{th} term of an A.P. is three times the first and the 7^{th} term exceeds twice the third term by 1. Find the first term and the common difference.

#### Solution

In the given problem, let us take the first term as *a* and the common difference as *d*

Here, we are given that,

`a_4 = 3a` ......(1)

`a_2 = 2a_3 + 1` .....(2)

We need to find a and d

So as we know

`a_n = a _ (n -1)d`

For the 4th term (n = 4)

`a_4 = a + (4 -1)d`

`3a = a + 3d`

3a - a = 3d

2a = 3d

`a = 3/2 d`

Similarly for the 3rd term (n = 3)

`a_3 = a + (3 - 1)d`

= a + 2d

Also for the 7th term (n = 7)

`a_7 = a + (7 - 1)d`

= a + 6d .....(3)

Now, using the value of *a*_{3} in equation (2), we get,

`a_7 = 2(a + 2d) + 1`

`= 2a + 4d + 1` ......(4)

Equating (3) and (4) we get

a + 6d = 2a + 4d +1

6d - 4d - 2a + a= +1

2d - a= +1

`2d - 3/2 d =1` `(a = 3/2 d)`

On further simplification, we get,

`(4d - 3d)/2 = 1`

d = (1)(2)

d = 2

Now to find a

`a = 3/2 d

`a =3/2 (2)`

a= 3

Therefore for the given A.P d= 2, a = 3