polynomials - Quadratic equation including Arithmetic Progression

For $a, b, c$ are real. Let $\frac{a+b}{1-ab}, b, \frac{b+c}{1-bc}$ be in arithmetic progression . If $\alpha, \beta$ are roots of equation $2acx^2+2abcx+(a+c) =0$ then find the value of $(1+\alpha)(1+\beta)$

Answer

The A.P. condition gives $2abc=a+c$ on simplifying, so the quadratic is effectively $x^2+bx+b=0$. As $(1+\alpha)(1+\beta) = 1+(\alpha+\beta) + (\alpha \beta)$, using Vieta we have this equal to $1+(-b)+b=1$.