Let a₁, a₂, a₃, be an arithmetic progression. It is given that a₅ = 8 and the product a₁ a₃ is minimum. If the sum of the first n terms of this progression is 21 , then the value of a_ n+2 + S_ n-1 is equal to

Step-by-step solution

Let the first term of the A.P. be a and the common difference be d . Given a₅ = 8 a + 4d = 8 a = 8 - 4d The product to be minimized is P = a₁ a₃ = a(a + 2d) . Substituting a = 8 - 4d into P : P = (8 - 4d)(8 - 2d) P = 8d^2 - 48d + 64 For P to be minimum, dP dd = 16d - 48 = 0 d = 3 . So, a = 8 - 4(3) = -4 . The sum of the first n terms is given as S_n = 21 . S_n = n 2 [2a + (n-1)d] = 21 n 2 [-8 + (n-1)3] = 21 n(3n - 11) = 42 3n^2 - 11n - 42 = 0 (n - 6)(3n + 7) = 0 Since n must be a positive integer, n = 6 . We need t