Let a₁, a₂, be an arithmetic progression of positive integers. If the sum of the first 12 terms is 468 and the 10^ th term satisfies 50 < a₁₀ < 55 , then the sum of the first 8 terms of the progression is equal to

Step-by-step solution

Let the first term of the arithmetic progression be a and the common difference be d . Since the terms are positive integers, a, d Z ^+ . Given the sum of the first 12 terms is 468 : S₁₂ = 12 2 (2a + 11d) = 468 6(2a + 11d) = 468 2a + 11d = 78 Since a 1 , we have 11d 76 d 6 . Also, 11d = 78 - 2a , which means 11d must be an even number. Thus, d must be an even positive integer. The possible values for d are 2, 4, and 6 . Case 1: If d = 2 , then 2a = 78 - 22 = 56 a = 28 . The 10^ th term is a₁₀ = a + 9d = 28 + 18 = 4