Question

Assertion (A) : If $\frac{a}{a_{1}}, \frac{b}{b_{1}}, \frac{c}{c_{1}}$ are in A.P., then $a_{1}, b_{1}, c_{1}$ are in G.P.
Reason (R) : If $a x^{2}+b x+c=0$ and $a_{1} x^{2}+b_{1} x+$ $c_{1}=0$ have a common root and $\frac{a}{a_{1}}, \frac{b}{b_{1}}, \frac{c}{c_{1}}$ are in A.P., then $2 a_{1}, b_{1}, 2 c_{1}$ are in G.P.

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A)
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A)
• C. (A) is true but (R) is false
• D. (A) is false but (R) is true

Answer 1

Answer: (D)

The equations $a x^{2}+b x+c=0$
and $a_{1} x^{2}+b_{1} x+$ $c_{1}=0$
have a common root then
$\left(a_{1} c-c_{1} a\right)^{2}=\left(a b_{1}-b a_{1}\right)\left(b c_{1}-c b_{1}\right) \,...(A)$
If $\frac{a}{a_{1}}, \frac{b}{b_{1}}, \frac{c}{c_{1}}$ are in A.P.,
then $\frac{a_{1} b-b_{1} a}{a_{1} b_{1}}=\frac{c b_{1}-b c_{1}}{b_{1} c_{1}}=d$
and $\frac{c a_{1}-a c_{1}}{a_{1} c_{1}}=2 d$
$\therefore $ From equation (A), we have,
$4 d^{2}\left(a_{1} c_{1}\right)^{2}=d^{2} a_{1} c_{1} b_{1}^{2}$
$\Rightarrow 4 a_{1} c_{1}=b_{1}^{2}$
$ \Rightarrow 2 a_{1}, b_{1}, 2 c_{1} \in G . P$
Hence Assertion (A) is false & Reason (R) is true