Given series is S = 42+39+36+...
Let first term of the sequence is a1 = a = 42.
Common difference of the sequence is d = a2 - a1 = 39 - 42 = -3.
∴ Sum of n terms of the sequence is Sn = n/2 [2a+(n-1)d]
Given that sum = 315
∴ n/2 [2 x 42 + (n-1) x -3] = 315
⇒ 42n -3/2 n(n-1) = 315
⇒ 84n - 3n2 + 3n = 630
⇒ 3n2 - 87n + 630 = 0
⇒ n2 - 29n + 210 = 0
⇒ n2 - 14n - 15n + 210 = 0
⇒ (n - 14) (n - 15) = 0
⇒ n = 14 or n = 15
Hence, sum of 14th terms of the series is 315. Also, sum of 15th terms is 315 (∵ a15 = a+14d = 0)