Question

The 12th term of an A.P. is 14 more than the 5th term. The sum of the first three terms is 36. Find the A.P.

Answer 1

The general term of an A.P. is an = a + (n − 1)d.

The problem states the 12th term is 14 more than the 5th term:

a₁₂ = a₅ + 14

(a + 11d) = (a + 4d) + 14

Subtract a from both sides and simplify:

11d − 4d = 14

7d = 14

d = `14/7`

d = 2

The sum of the first three terms is 36:

a₁ + a₂ + a₃ = 36

a + (a + d) + (a + 2d) = 36

3a + 3d = 36

Substitute d = 2

3a + 3(2) = 36

3a + 6 = 36

⇒ 3a = 30

⇒ a = `30/3`

⇒ a = 10

a₂ = a + (2 − 1)2

= 10 + 2

= 12

a₃ = a + (3 − 1)2

= 10 + 4

= 14

a₄ = a + (4 − 1)2

= 10 + 6

= 16

a₅ = a + (4 − 1)2

= 10 + 6

= 16