Let a₁, a₂, a₃, be an arithmetic progression with a non-zero common difference, and let b₁, b₂, b₃, be a geometric progression with a positive common ratio. If a₁ = b₁ = 2 , a₂ = b₂ , and a₆ = b₃ , then the value of a₄ + b₄ is equal to

Step-by-step solution

Let the common difference of the A.P. be d ( d 0 ) and the common ratio of the G.P. be r ( r > 0 ). Given a₁ = b₁ = 2 . Since a₂ = b₂ , we have: a₁ + d = b₁ r 2 + d = 2r d = 2r - 2 Also given a₆ = b₃ , so: a₁ + 5d = b₁ r^2 2 + 5(2r - 2) = 2r^2 2 + 10r - 10 = 2r^2 2r^2 - 10r + 8 = 0 r^2 - 5r + 4 = 0 (r - 1)(r - 4) = 0 Thus, r = 1 or r = 4 . If r = 1 , then d = 2(1) - 2 = 0 , which contradicts the given condition d 0 . Therefore, r = 4 . Substituting r = 4 back to find d : d = 2(4) - 2 = 6 Now, we need to find a₄ + b