Consider the quadratic equation (n^2 - 2n + 2 | vedclass

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(C) The product of the roots $P$ of the quadratic equation $ax^2 + bx + c = 0$ is given by $P = c/a$.Here,$P = \frac{n^2 - 2n + 2}{n^2 - 2n + 2} = 1$.

Vedclass · Accepted Answer

$364$/$243$

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(C) The product of the roots $P$ of the quadratic equation $ax^2 + bx + c = 0$ is given by $P = c/a$.Here,$P = \frac{n^2 - 2n + 2}{n^2 - 2n + 2} = 1$. Since the product is constant,its minimum value $\alpha = 1$.The sum of the roots $S$ is given by $S = -b/a = \frac{3}{n^2 - 2n + 2}$.To maximize $S$,we must minimize the denominator $n^2 - 2n + 2 = (n-1)^2 + 1$.The minimum value of the denominator is $1$ (at $n=1$),so the maximum value of the sum is $\beta = 3/1 = 3$.For the $G$.$P$.,the first term $a = \alpha = 1$ and the common ratio $r = \alpha/\beta = 1/3$.The sum of the first $n$ terms of a $G$.$P$. is $S_n = \frac{a(1-r^n)}{1-r}$.For $n=6$,$S_6 = \frac{1(1-(1/3)^6)}{1-1/3} = \frac{1 - 1/729}{2/3} = \frac{728/729}{2/3} = \frac{728}{729} 	imes \frac{3}{2} = \frac{364}{243}$.

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