The 7th term of an A.P. is 20 and its 13th term is 32. Find the A.P. [CBSE 20041

We have

a₇ = a + (7 – 1)d = a + 6d = 20 …(1)

and a₁₃ = a + (13 – 1)d = a + 12d = 32 …(2)

Solving the pair of linear equations (1) and (2), we get

a + 6d – a – 12d = 20 – 32

⇒ – 6d = –12

⇒ d = 2

Putting the value of d in eq (1), we get

a + 6(2) = 20

⇒ a + 12 = 20

⇒ a = 8

Hence, the required AP is 8, 10, 12, 14,…