If $a/a_1 , b/b_1 , c/c_1 $ are in A.P. then $ a_1 , b_1 , c_1 $ are in G.P.
I have no idea , how to approach this .
What I have thought :
For the AP series $ a/a_1 = k - d $
the rest be k & k + d respectively
The condition for common root here in this case is :
$$ (ca_1 - c_1a )^2 = ( ab_1 - a_1b ) ( bc_1 - b_1c ) $$
Replacing a , b , c with : $ ( k - d ) a_1 , k b_1 , ( k + d ) c_1 $ now thats too messy . Is there a better approach .
On
I thought it will be messy , but apparently its not .
I will proceed from $a = ( k - d ) a_1 , b = kb_1 , c = ( k + d )c_1$
Now in this case , applying the condition for a single common root in two equations :
$$ ( ca_1 - c_1a )^2 = ( 2bc_1 - 2b_1c ) ( 2ab_1 - 2a_1b ) $$
Replacing $ a,b,c $ with $ ( k - d ) a_1 , kb_1 , ( k + d )c_1 $
$$ \implies ( 2 c_1a_1d )^2 = ( -2a_1b_1d ) ( -2b_1c_1d ) $$ $$ \implies a_1c_1 = { b_1 }^2 $$
Let $a=a_1(k-d)$, $b=b_1k$ and $c=c_1(k+d)$ now $ax^2+2bx+c=0$ becomes
$$a_1(k-d)x^2+2b_1kx+c_1(k+d)=0$$ $\implies$
$$k(a_1x^2+2b_1x+c_1)-a_1dx^2+c_1d=0 \tag{1}$$ and
the other quadratic is $$a_1x^2+2b_1x+c_1=0 \tag{2}$$ If $(1)$ and $(2)$ have common root $\alpha$ we have
from $(1)$
$$k(0)-a_1d \alpha^2+c_1d=0$$ $\implies$
$$\alpha=\sqrt{\frac{c_1}{a_1}}$$ and substitute this in $(2)$ we get
$$a_1(\frac{c_1}{a_1})+2b_1\sqrt{\frac{c_1}{a_1}}+c_1=0$$ simplifying this we get $$b_1^2=a_1c_1$$