Let a₁, a₂, a₃, be an A.P. and g₁ = a₁, g₂, g₃, be an increasing G.P. If a₁ = a₂ + g₂ = 1 and a₃ + g₃ = 4 , then a₁₀ + g₅ is equal to:

Step-by-step solution

Let the common difference of the A.P. be d and the common ratio of the G.P. be r . Given a₁ = 1 and g₁ = a₁ = 1 . From a₂ + g₂ = 1 , we have: (a₁ + d) + g₁ r = 1 1 + d + r = 1 d = -r From a₃ + g₃ = 4 , we have: (a₁ + 2d) + g₁ r^2 = 4 1 + 2d + r^2 = 4 2d + r^2 = 3 Substituting d = -r into the equation: -2r + r^2 = 3 r^2 - 2r - 3 = 0 (r - 3)(r + 1) = 0 Since the G.P. is increasing and g₁ = 1 , the common ratio r must be greater than 1 . Thus, r = 3 . Then, d = -3 . We need to find a₁₀ + g₅ : a₁₀ = a₁ + 9d = 1 + 9(-3)