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 $a{x}^2+bx+c=0$ 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 $2a_1,b_1,2c_1$ 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 $a{x}^2+bx+c=0$
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 $\frac{c{a}_1-a{c}_1}{a_1{c}_1}=2d$
$\therefore$ From equation (A), we have,
$4d^2{\left(a_1{c}_1\right)}^2=d^2{a}_1{c}_1{b}_1^2$
$\Rightarrow 4a_1{c}_1=b_1^2$
$\Rightarrow 2a_1,b_1,2c_1\in G.P$
Hence Assertion (A) is false & Reason (R) is true