Question

Assertion :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: If $a{x}^2+bx+c$ and $a_1{x}^2+b_1{x}^2+c_1=0$ have a common root & $\frac{a}{a_1},\frac{b}{b_1},\frac{c}{c_1}$ are in $A.P$ then $a_1,b_1,c_1$ are in $G.P$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect, Reason is correct

Answer 1

The correct option is D Assertion is incorrect, Reason is correct

$a{x}^2+bx+c=0$

Formula for $a_1{x}^2+b_{1}x+c_1=0$ and $a_2{x}^2+b_{2}x+c_2=0$ to have common roots is

${(a}_1{b}_2-a_2{b}_1\left)\right(b_1{c}_2-b_2{c}_1)={(c_1{a}_2-c_2{a}_1)}^2$

So, In our case we have $(a{b}_1-a_{1}b\left)\right(b{c}_1-b_{1}c)={(c{a}_1-c_{1}a)}^2$

Rearranging the terms gives:

$a_1{b}_1(\frac{a}{a_1}-\frac{b}{b_1})c_1{b}_1(\frac{b}{b_1}-\frac{c}{c_1})={a_1}^2{c_1}^2{(\frac{c}{c_1}-\frac{a}{a_1})}^2$

As $\frac{a}{a_1},\frac{b}{b_1},\frac{c}{c_1}$ are in AP

Let $d$ be the common difference

$\frac{a}{a_1}-\frac{b}{b_1}=-d,\frac{b}{b_1}-\frac{c}{c_1}=-d,\frac{c}{c_1}-\frac{a}{a_1}=2d$

$a_1{b}_1(-d)c_1{b}_1(-d)={a_1}^2{c_1}^2\left(4d^2\right)$

${b_1}^2=4a_1{c}_1$

So $a_1,b_1,c_1$ are not in GP

Assertion is incorrect and reason is also incorrect.