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 $$(1+\alpha)(1+\beta)$$

• What have you tried? – Yuta May 2 at 13:30
• What means "are in ap."? – Dietrich Burde May 2 at 13:31
• Yes i have tried that if abc are in ap then 2b=a+c. I tried to get a relation between a b and c. But i got to nowhere – Suman Chandra May 2 at 13:31
• ap is arithmetic progression – Suman Chandra May 2 at 13:32
• @Suman Chandra So $\frac{a+b}{1-ab}$, $b$ and $\frac{b+c}{1-bc}$ are in ap gives $\frac{a+b}{1-ab}+\frac{b+c}{1-bc}=2b$. This condition may be useful. – Yuta May 2 at 13:34

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 $$1+(-b)+b=1$$.
Hint: $$(a+b)(1-ac)+(b+c)(1-ab)=2b(1-ab)(1-ac)$$ gives us $$-(1+b^2)(-a-c+2abc)=0$$,so it must be $$a+c=2abc$$ Now we must solve the quadratic $$x_{1,2}=-\frac{b}{2}\pm\sqrt{\frac{b^2}{4}-\frac{a+c}{2ac}}$$ Plugging the term $$a+c=2abc$$ into the solution of the quadric equation we get $$x_{1,2}=-\frac{b}{2}\pm\sqrt{\frac{b^2}{4}-b}$$