Consider the quadratic equation . Let be the minimum value of the product of its roots and 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 and the common ratio is , is: [2026]

Question

Consider the quadratic equation $(n^{2} - 2 n + 2) x^{2} - 3 x + {(n^{2} - 2 n + 2)}^{2} = 0, n \in R$ . Let $α$ be the minimum value of the product of its roots and $β$ 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 $α$ and the common ratio is $\frac{α}{β}$ , is: [2026]

Smart Achievers · Accepted Answer

(3)

Given, $(n^{2} - 2 n + 2) x^{2} - 3 x + {(n^{2} - 2 n + 2)}^{2} = 0$

Let $k = n^{2} - 2 n + 2$

$\Rightarrow k = {(n - 1)}^{2} + 1$

Min value of k = 1 (at n = 1)

Thus, $k x^{2} - 3 x + k^{2} = 0$

Sum of roots $= - \frac{(- 3)}{k} = \frac{3}{k}$

$\Rightarrow$ Maximum value of sum of roots $= \frac{3}{1} = 3$

$\Rightarrow β = 3$

Product of roots $= \frac{k^{2}}{k} = k$ .

$\Rightarrow$ Minimum value of product of roots = 1

$\Rightarrow α = 1$

Since, $α = 1$

$∴$ Common ratio $(r) = \frac{α}{β} = \frac{1}{3}$

$∵ S_{n} = \frac{a (1 - r^{n})}{1 - r}$

$∴ S_{6} = (\frac{1 (1 - {(\frac{1}{3})}^{6})}{1 - \frac{1}{3}})$

$\Rightarrow S_{6} = \frac{1 - \frac{1}{729}}{\frac{2}{3}} = \frac{364}{243}$

Solution

1 $\frac{61}{37}$

2 $\frac{121}{81}$

3 $\frac{364}{243}$

4 $\frac{1093}{729}$

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Solution

Q. Consider the quadratic equation (n2–2n+2)x2–3x+(n2–2n+2)2=0, n∈R. Let α be the minimum value of the product of its roots and β 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 α and the common ratio is αβ, is: [2026]

1 6137

2 12181

3 364243

4 1093729

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1 $\frac{61}{37}$

2 $\frac{121}{81}$

3 $\frac{364}{243}$

4 $\frac{1093}{729}$