December 27, 2024
Explanation:
(b)
Sum of first ‘n’ term of an AP,
Sn = n/2 [2a + (n−1)d]
Sum of first 15 terms =15/2 [2a+14d] = 200
⇒ a + 7d = 40/3 -----(I)
Sum of next 15 terms = Sum of first 30 terms – Sum of first 15 terms
= 30/2 [2a + 29d] −15/2 [2a+14d] = 350
⇒ 4a + 58d − 2a − 14d =140/3
⇒ 2a + 44d = 140/3 ⇒ a + 32d =70/3 --------(2)
Solving equation (1) and equation (2), we get d = 2/3