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
$\left(1-ab\right)m^{2} -\left(a^{2}+b^{2}\right)m-\left(1+ab\right)=0$
$\Rightarrow m\left(a^{2}+b^{2}\right)+\left(m^{2}+1\right) ab = m^{2}-1\quad...\left(1\right)$
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)$