How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

Here, a = 9, d = 17 – 9 = 8 and S_n = 636.
The sum of n terms of an AP is given by
S_n = n/2[2a + (n -1)d]
⇒ 636 = n/2[2(9) + (n -1)(8)]
⇒ 636 = n[9 + 4n – 4]
⇒ 4n² + 5n – 636 = 0
⇒ 4n² + 53n – 48n – 636 = 0
⇒ n(4n + 53) – 12(4n + 53) = 0
⇒ (n – 12)(4n + 53) = 0
⇒ n – 12 = 0 [∵ sn + 53 ≠ 0 as n ≠ – 53/4]
⇒ n = 12
Hence, 12 terms of the AP: 9, 17, 25 … must be taken to get the sum 636.