Question
Consider the quadratic equation $\left( n^{2} - 2n + 2 \right)x^{2} - 3x + \left( n^{2} - 2n + 2 \right)^{2} = 0,n \in R$. Let $\alpha$ be the minimum value of the product of its roots and $\beta$ be the maximum value of the sum of its roots. Then the sum of the first six terms of the G.P., whose first term is $\alpha$ and the common ratio is $\frac{\alpha}{\beta}$, is :
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Solution
Sol. [Key Concept: $\alpha = (n{- 1)}^{2} + 1$ has minimum $1$ and $\beta = \frac{3}{(n{- 1)}^{2} + 1}$ has maximum $3$; sum six terms of the G.P. with first term $\alpha$ and ratio $\frac{\alpha}{\beta}$.]
$${\alpha = n^{2} - 2n + 2 }{\alpha = (n - 1)^{2} + 1 }$$∴ minimum value of $\alpha$ is 1
Similarly $\beta = \frac{3}{(n - 1)^{2} + 1}$
∴ maximum value of $\beta$ is 3
required G.P. is $1,\frac{1}{3},\frac{1}{3^{2}},\ldots$
$$S_{6} = \frac{1\left( 1 - \left( \frac{1}{3} \right)^{6} \right)}{1 - \frac{1}{3}} = \frac{3}{2}\left\lbrack 1 - \left( \frac{1}{3} \right)^{6} \right\rbrack = \frac{364}{243} $$
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