Question

The 4th term of an AP is three times the first and the 7th term exceeds twice the third term by 1. Find the first term 'a' and common difference 'd'.​

Answer 1

a+3d= 3a

-2a+3d=0

Now

a+6d=2(a+2d)+1

-a-4d=2

Now compare eqn frst and second

We will get

-2a+3d=0

a+4d=-2

Then d =4/7

And a= -30/7

Answer 2

There is an AP, The 4th term is three times the first. So the relation is

3a = a + 3d, or

2a = 3d, or

a = (3/2)d …(1)

The 7th term exceeds twice the third term by 1. The relation is

a+6d = 2(a+2d)+1, or

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

2a-a = 6d-4d-1, or

a = 2d-1 …(2). Equate (1) and (2)

a = (3/2)d = 2d-1, or

3d = 4d-2, or

4d-3d=2, or d = 2 and a = 3.

The first term is 3 and the common difference is 3.

Check: The series of the AP is 3,5,7,9,11,13,15.

The 4th term is 9, which is three times the first term. Correct.

The 7th term is 15, which is twice the third term plus 1. Correct.