# If positive numbers a^-1, b^-1, c^-1 are in AP, then the product of the roots of the equation x^2-kx+2b^101-a^101-c^101=0, (k is real) is greater than 0, less than 0 or equal to 0?

Jul 22, 2017

See below.

#### Explanation:

If ${a}^{-} 1 , {b}^{-} 1 , {c}^{-} 1$ are in AP then

${b}^{-} 1 = {a}^{-} 1 + r$ and ${c}^{-} 1 = {a}^{-} 1 + 2 r$ or

$b = \frac{a}{a r + 1}$ and $c = \frac{a}{2 a r + 1}$

Now, regarding the the product of the roots of

${x}^{2} - k x + 2 {b}^{101} - {a}^{101} - {c}^{101} = 0$ this product is given by

$2 {b}^{101} - {a}^{101} - {c}^{101}$ or

$2 {\left(\frac{a}{a r + 1}\right)}^{101} - {a}^{101} - {\left(\frac{a}{2 a r + 1}\right)}^{101}$ or

${a}^{101} \left(2 {\left(\frac{1}{a r + 1}\right)}^{101} - {\left(\frac{1}{2 a r + 1}\right)}^{101} - 1\right)$ but

$2 {\left(\frac{1}{a r + 1}\right)}^{101} - {\left(\frac{1}{2 a r + 1}\right)}^{101} < 1$ then

$2 {\left(\frac{1}{a r + 1}\right)}^{101} - {\left(\frac{1}{2 a r + 1}\right)}^{101} - 1 < 0$

NOTE:

$\frac{2}{a r + 1} - \frac{1}{2 a r + 1} = \frac{3 a r}{3 a r + 1 + 2 {a}^{2} {r}^{2}} < 1$