The roots of the equation

x^5 - 40x^4 + px | vedclass

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Question

$a+a r+a r^{2}+a r^{3}+a r^{4}=40$

$\frac{1}{a}+\frac{1}{a r}+\frac{1}{a r^{2}}+\frac{1}{a r^{3}}+\frac{1}{a r^{4}}=10$

$\left(a r^{2}\right)^{2

Vedclass · Accepted Answer

$32$

Vedclass · Answer

$a+a r+a r^{2}+a r^{3}+a r^{4}=40$

$\frac{1}{a}+\frac{1}{a r}+\frac{1}{a r^{2}}+\frac{1}{a r^{3}}+\frac{1}{a r^{4}}=10$

$\left(a r^{2}\right)^{2}=4 \Rightarrow a r^{2}=\pm 2$

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 $\left| s \right|$ is

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