Clean Minima

$\begin{aligned} & \alpha +\beta =4({{k}^{2}}-7k+8) \\ & \alpha \beta =-7 \\ & \frac{{{V}_{101}}-7{{V}_{99}}}{{{V}_{100}}} \\ & \frac{{{V}_{101}}+(\alpha \beta ){{V}_{99}}}{{{V}_{100}}} \\ & \frac{{{\alpha }^{101}}-{{\beta }^{101}}+{{\alpha }^{100}}\beta -\alpha {{\beta }^{100}}}{{{\alpha }^{100}}-{{\beta }^{100}}} \\ & \frac{{{\alpha }^{101}}+{{\alpha }^{100}}\beta -({{\beta }^{101}}+\alpha {{\beta }^{100}})}{{{\alpha }^{100}}-{{\beta }^{100}}} \\ & \frac{{{\alpha }^{100}}(\alpha +\beta )-{{\beta }^{100}}(\alpha +\beta )}{{{\alpha }^{100}}-{{\beta }^{100}}} \\ & \frac{({{\alpha }^{100}}-{{\beta }^{100}})(\alpha +\beta )}{{{\alpha }^{100}}-{{\beta }^{100}}} \\ & \alpha +\beta =4({{k}^{2}}-7k+8) \\ \end{aligned}$ The Minimum Value of $4({{k}^{2}}-7k+8)$ is -17 So, the solution is -17

Let $M$ denote the expression $\large \frac {V_{101} - 7V_{99} }{V_{100}}$ , rearranging gives

$V_{101} - M \cdot V_{100} - 7 V_{99} = 0$

By Newton's sum, comparing it to the quadratic equation given:

$M = 4(k^2 - 7k + 8)$

The minimum value of $M$ occur when $\frac {\mathrm d M}{\mathrm d k} = 0$ , which means $4(2k-7) = 0 \Rightarrow k = \frac {7}{2}$ , substitute this value into $M$ yields $\boxed{-17}$

similar type was asked in jee mains 2015