Question

If $m$ is a root of the equation $(1-ab)x^2-(a^2+b^2)x-(1+ab)=0$, and $m$ harmonic means are inserted between $a$ and $b$, then the difference between the last and the first of the means equals

• A. $b-a$
• B. $ab(b-a)$
• C. $a(b-a)$
• D. $ab(a-b)$

Answer 1

Answer: (B)

By the given condition
$\left(1-ab\right)m^2-\left(a^2+b^2\right)m-\left(1+ab\right)=0$
$\Rightarrow m\left(a^2+b^2\right)+\left(m^2+1\right)ab=m^2-1...\left(1\right)$
Now $H_1=$ First $H.M$. between $a$ and $b$
$=\frac{\left(m+1\right)ab}{a+mb}$ and $H_m=\frac{\left(m+1\right)ab}{b+ma}$
$\therefore H_m-H_1=\left(m+1\right)ab\left[\frac{1}{b+ma}-\frac{1}{a+mb}\right]$
$=\left(m+1\right)ab\frac{\left[\left(m-1\right)\left(b-a\right)\right]}{\left(b+ma\right)\left(a+mb\right)}$ $=\frac{\left(m^2-1\right)ab\left(b-a\right)}{m\left(a^2+b^2\right)+\left(m^2+1\right)ab}$
$=\frac{\left(m^2-1\right)ab\left(b-a\right)}{m^2-1}$ [By $(1)]$
$=ab\left(b-a\right)$