<s>[INST] An Expansion Problem (Algebra) In the expansion of $(1+x)^n$, if the Coeff of $T_r$ : Coeff of $T_{r+1}$ : Coeff of $T_{r+2}$ = a:b:c, prove that $r = \frac{a(b+c)}{b^2-ac}$, where $b^2$ does not equal ac, and a, b, c $\in Z^+$
 [/INST] We know that the coefficient of the $r$th term in the expansion of $(1+x)^n$ is : $\binom{n}{r-1}$. Similarly the coefficients of the $(r+1)$th and $(r+2)$th term are respectively: $\binom{n}{r} \text{and} \binom{n}{r+1}$.   
Thus, we have, $$\frac{T_{r+1}}{T_r} =\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r} =\frac{b}{a} ...(1)$$ $$\frac{T_{r+2}}{T_{r+1}} =\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{n-r}{r+1} = \frac{c}{b} ...(2)$$ Then we have from $(1)$ and $(2)$, $$n = \frac{b}{a}r-(-r+1) = \frac{c}{b}(r+1) +r$$ Solving for $r$, we thus get,$$\boxed{r =\frac{a(b+c)}{b^2-ac}}$$ Hope it helps.
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