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 and common difference.

Answer 1

Let a be the first term and d be the common difference.

Given that 4th term of an AP is three times the first.

T4 = 3T1

a + (4 - 1) * d = 3 * a

a + 3d = 3a.

3d = 3a - a

3d = 2a. 

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


Given that 7th term exceeds twice the third term by 1.

T7 = 2T3 + 1

a + (7 - 1) *d = 2(a + (3 - 1) * d)+ 1

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

a = 2d - 1

3d/2 = 2d - 1

3d = 4d - 2

3d - 4d = -2

-d = -2

d = 2.


Substitute d = 2 in (1), we get

a = 3 * 2/2

   = 3.

Therefore the first term = 3 and the common difference = 2.

The series is 3,5,7,9...


Hope this helps!

Answer 2

Answer:

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