# 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,…

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