Let a₁, a₂, a₃, be an arithmetic progression with common difference d 0 and b₁, b₂, b₃, be a geometric progression with common ratio r > 1 . Both progressions have the same first term , where is a positive integer. If a₂ = b₂ and a₄ = b₃ , and the minimum value of the quadratic polynomial f(x) = x^2 - a₃ x + b₄ is -4 , then the value of a₅ + b₅ is equal to

Step-by-step solution

Given a₁ = b₁ = . From a₂ = b₂ , we have + d = r d = (r - 1) . From a₄ = b₃ , we have + 3d = r^2 . Substituting d = (r - 1) into the second equation: + 3 (r - 1) = r^2 Since 0 , dividing by gives: 1 + 3r - 3 = r^2 r^2 - 3r + 2 = 0 (r - 1)(r - 2) = 0 Since r > 1 , we must have r = 2 . Then d = (2 - 1) = . Now, we find a₃ and b₄ in terms of : a₃ = + 2d = + 2 = 3 b₄ = r^3 = (2)^3 = 8 The given polynomial is f(x) = x^2 - a₃ x + b₄ = x^2 - 3 x + 8 . The minimum value of a quadratic Ax^2 + Bx + C occurs at x = - B 2A and