Question

Given that $\alpha ,\gamma$ are roots of the equation $A{x}^2-4x+1=0$ and $\beta ,\sigma$ the roots of the equation $B{x}^2-6x+1=0$, then find the value of (A+B), such that $\alpha ,\beta ,\gamma$ & $\sigma$ are in H.P.

Answer 1

Let $\alpha =\frac{1}{a}$, $\beta =\frac{1}{a+d}$, $\gamma =\frac{1}{a+2d}$, $\sigma =\frac{1}{a+3d}$

So, $\alpha +\gamma =\frac{1}{a}+\frac{1}{a+2d}=\frac{2(a+d)}{a(a+2d)}=\frac{2\alpha \gamma }{\beta }=\frac{4}{A}$

$\Rightarrow \frac{\alpha \gamma }{\beta }=\frac{2}{A}$

Similarly, $\frac{\beta \sigma }{\gamma }=\frac{3}{A}$

But, $\alpha \gamma =\frac{1}{A}$ and $\beta \sigma =\frac{1}{B}$

ie, $\beta =\frac{1}{2}$ and $\gamma =\frac{1}{3}$

As all of them are in Harmonic Progression,

$\alpha =1$ and $\sigma =\frac{1}{4}$

Then, $A+B=\frac{1}{\alpha \gamma }+\frac{1}{\beta \sigma }=3+8=11$