How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
Let there be n terms of this A.P.
For this A.P., a = 9
d = a2 − a1 = 17 − 9 = 8
As Sn = n/2 [2a + (n - 1)d]
636 = n/2 [2 × a + (8 - 1) × 8]
636 = n/2 [18 + (n- 1) × 8]
636 = n [9 + 4n − 4]
636 = n (4n + 5)
4n2 + 5n − 636 = 0
4n2 + 53n − 48n − 636 = 0
n (4n + 53) − 12 (4n + 53) = 0
(4n + 53) (n − 12) = 0
Either 4n + 53 = 0 or n − 12 = 0
n = (-53/4) or n = 12
n cannot be (-53/4). As the number of terms can neither be negative nor fractional, therefore, n = 12 only.
