In the binomial expansion of (ax + b x^2 )¹⁴ , where a and b are non-zero real numbers, the coefficients of x^2 , x⁻¹ , and x⁻⁴ are in Arithmetic Progression in that order. The sum of all possible values of a b is:

Step-by-step solution

The general term in the expansion of (ax + b x^2 )¹⁴ is given by: T_ r+1 = ¹⁴C_r (ax)^ 14-r ( b x^2 )^r = ¹⁴C_r a^ 14-r b^r x^ 14-3r We need the coefficients of x^2 , x⁻¹ , and x⁻⁴ . For x^2 : 14 - 3r = 2 r = 4 . The coefficient is ¹⁴C₄ a¹⁰ b^4 . For x⁻¹ : 14 - 3r = -1 r = 5 . The coefficient is ¹⁴C₅ a^9 b^5 . For x⁻⁴ : 14 - 3r = -4 r = 6 . The coefficient is ¹⁴C₆ a^8 b^6 . Since these coefficients are in Arithmetic Progression, we have: 2 (¹⁴C₅ a^9 b^5) = ¹⁴C₄ a¹⁰ b^4 + ¹⁴C₆ a^8 b^6 Dividing the entire equation by