The roots of the equation x^5 - 40x^4 + px^3 | vedclass

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(D) Let the roots be $\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$. From the sum of roots,$\frac{a}{r^2} + \frac{a}{r} + a + ar + ar^2 = 40$. From the sum

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$32$

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(D) Let the roots be $\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$. From the sum of roots,$\frac{a}{r^2} + \frac{a}{r} + a + ar + ar^2 = 40$. From the sum of reciprocals,$\frac{r^2}{a} + \frac{r}{a} + \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} = 10$. This can be written as $\frac{1}{a} (r^2 + r + 1 + \frac{1}{r} + \frac{1}{r^2}) = 10$. Also,the sum of roots is $a(\frac{1}{r^2} + \frac{1}{r} + 1 + r + r^2) = 40$. Dividing the two equations: $\frac{a(r^2 + r + 1 + \frac{1}{r} + \frac{1}{r^2})}{\frac{1}{a}(r^2 + r + 1 + \frac{1}{r} + \frac{1}{r^2})} = \frac{40}{10}$ $\Rightarrow a^2 = 4$ $\Rightarrow a = 2$ (assuming positive roots). The product of the roots is $s = -(\frac{a}{r^2} \cdot \frac{a}{r} \cdot a \cdot ar \cdot ar^2) = -a^5$. Since $a = 2$,$s = -(2)^5 = -32$. Therefore,$|s| = |-32| = 32$.

The roots of the equation $x^5 - 40x^4 + px^3 + qx^2 + rx + s = 0$ are in $G.P.$ The sum of their reciprocals is $10$. Then the value of $|s|$ is

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