Let $$a_{1} a_{2}, a_{3}$$ be three distinct real numbers in geometric progression. If the equations $$a_{1}x^{2} + 2a_{2} x + a_{3} = 0$$ and $$b_{1}x^{2} + 2b_{2}x + b_{3} = 0$$ has a common root,then which of the following is necessarily true?

Given, $$a_{1} a_{2}, a_{3}$$ are in G.P.

Let the common ratio of the G.P. be $$r$$

So, $$a_2=a_1r,\ a_3=a_1r^2$$

Putting the values in equation, $$a_{1}x^{2} + 2a_{2} x + a_{3} = 0$$

So, $$a_1x^2+2a_1xr+a_1r^2=0$$

or, $$a_1\left(x^2+2xr+r^2\right)=0$$

or, $$a_1\left(x+r\right)^2=0$$

Now, $$a_1\ne\ 0$$

So, $$\left(x+r\right)^2=0$$

or, $$x=-r$$ is the only root

So the common root is also $$x=-r$$

Putting this in the second equation, $$b_{1}x^{2} + 2b_{2}x + b_{3} = 0$$

or, $$b_{1}r^{2} - 2b_{2}r + b_{3} = 0$$

Now we can replace $$r^2=\dfrac{a_3}{a_1},\ r=\dfrac{a_2}{a_1}$$

So, $$b_1\cdot\dfrac{a_3}{a_1}+b_3=2b_2\cdot\dfrac{a_2}{a_1}$$

Dividing all terms by $$a_3$$,

$$\dfrac{b_1}{a_1}+\dfrac{b_3}{a_3}=2\cdot\dfrac{b_2a_2}{a_1a_3}=\dfrac{2b_2a_2}{a_2^2}=\dfrac{2b_2}{a_2}$$

(Since, $$a_1,a_2,a_3$$ are in G.P., $$a_1\cdot a_3=a_2^2$$)

so, $$\dfrac{b_1}{a_1}+\dfrac{b_3}{a_3}=\dfrac{2b_2}{a_2}$$

or, $$\dfrac{b_1}{a_1},\dfrac{b_2}{a_2},\dfrac{b_3}{a_3}$$ are in A.P.