Let S_n denote the sum of the first n terms of an arithmetic progression whose terms are all integers. If S₈ = 120 , S₄ S₃ - 1 and S₁₁ < S₁₀ + 245 , then the sum of all possible values of the common difference of this progression is

Step-by-step solution

Let the first term of the arithmetic progression be a and the common difference be d . Since all terms are integers, both a and d must be integers. Given S₈ = 120 : 8 2 [2a + (8 - 1)d] = 120 4(2a + 7d) = 120 2a + 7d = 30 2a = 30 - 7d Since a is an integer, 2a is an even integer. The number 30 is even, so 7d must also be even, which implies that d must be an even integer. From the first inequality: S₄ S₃ - 1 S₄ - S₃ -1 T₄ -1 a + 3d -1 Multiplying by 2: 2a + 6d -2 Substituting 2a = 30 - 7d : (30 - 7d) + 6d -2 30 - d