The first term of an arithmetic progression is 13 . If the sum of its first 4 terms is 5 times the sum of its next 4 terms, then the maximum possible sum of the first n terms of this progression is

Step-by-step solution

Let a be the first term and d be the common difference. We are given a = 13 . The sum of the first 4 terms is 5 times the sum of the next 4 terms: S₄ = 5(S₈ - S₄) 6S₄ = 5S₈ Using the sum formula S_n = n 2 [2a + (n-1)d] : 6 4 2 [2(13) + 3d] = 5 8 2 [2(13) + 7d] 12[26 + 3d] = 20[26 + 7d] 3[26 + 3d] = 5[26 + 7d] 78 + 9d = 130 + 35d 26d = -52 d = -2 The sequence is 13, 11, 9, To find the maximum sum of the first n terms, we must add all the non-negative terms. The n^ th term is: T_n = a + (n-1)d = 13 + (n-1)(-2) = 15 -