If 10th term of an A.P. is 52 and 17th term is 20 more than the 13th term, find the A.P.

Given: a₁₀ = 52 and a₁₇ = 20 + a₁₃

Now, a_n = a + (n – 1)d

a₁₀ = a + (10 – 1)d

52 = a + 9d …(i)

and a₁₇ = 20 + a₁₃

a + (17 – 1)d = 20 + a + (13 – 1)d

⇒ a + 16d = 20 + a + 12d

⇒ 16d –12d = 20

⇒ 4d = 20

⇒ d = 5

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

a + 9(5) = 52

⇒ a + 45 = 52

⇒ a = 52 – 45

⇒ a = 7

Therefore, the AP is 7, 12, 17, …