In an Ap of 50 terms the sum of first 10 terms is 210 and sum of its 15 terms is 2565.find the ap
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Let a be the first term and d be the common difference of the given AP.Sum of the first n terms is given bySn = n/2 {2a + (n – 1)d}Putting n = 10, we getS₁₀ = 10/2 {2a + (10 – 1)d}210 = 5 (2a + 9d) 2a + 9d = 210/52a + 9d = 42 ……………(1)Sum of the last 15 terms is 2565⇒ Sum of the first 50 terms – sum of the first 35 terms = 2565S₅₀ – S₃₅ = 2565⇒ 50/2 {2a + (50 – 1)d} – 35/2 {2a + (35 – 1)d} = 256525 (2a + 49d) – 35/2 (2a + 34d) = 2565⇒ 5 (2a + 49d) – 7/2 (2a + 34d) = 513⇒ 10a + 245d – 7a + 119d = 513⇒ 3a + 126d = 513 ⇒ a + 42d = 171 ……..(2)Multiply the equation (2) with 2, we get2a + 84d = 342 ………(3)Subtracting (1) from (3) 2a + 84d = 342 2a + 9d = 42- – -_______________ 75d = 300_______________ d= 4Now, substituting the value of d in equation (1)2a + 9d = 422a + 9*4 = 422a = 42 – 362a = 6a = 3So, the required AP is 3, 7, 11, 15, 19, 23, 27, 31, 35, 39