Question

The 4 term of ap is equal to 3times the first term and the 7 th term exceeds twice the 3rd term by 1 find ap

Answer 1

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.

please mark as BRAINliest answer

Answer 2

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

⇒3a=a+3d

⇒3a-a=3d

⇒2a=3d

⇒a=(3/2)d.......(1)

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

⇒a+6d=2(a+2d)+1

⇒a+6d=2a+4d+1

⇒6d-4d-1=2a-a

⇒2d-1=a........(2)

By Equation 1&2

a=(3/2)d=2d-1                       {by cross multiply}

=3d=4d-2

=2=4d-3d

=2=d

So, d= 2

and put the value of d in equation 2

a⇒2(2)-1

a⇒4-1

a⇒3

And a=3, d=2

So, we can write the AP is 3,5,7,9,11,13,15