If m is a root of the equation (1 - ab)x^2 - | vedclass

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(B) Given the equation $(1 - ab)m^2 - (a^2 + b^2)m - (1 + ab) = 0$.This implies $m(a^2 + b^2) + (m^2 + 1)ab = m^2 - 1$ ......$(i)$Let $H_1, H_2, \dots

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$ab(b - a)$

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(B) Given the equation $(1 - ab)m^2 - (a^2 + b^2)m - (1 + ab) = 0$.This implies $m(a^2 + b^2) + (m^2 + 1)ab = m^2 - 1$ ......$(i)$Let $H_1, H_2, \dots, H_m$ be the $m$ harmonic means between $a$ and $b$.The first harmonic mean $H_1 = \frac{(m + 1)ab}{a + mb}$ and the last harmonic mean $H_m = \frac{(m + 1)ab}{b + ma}$.The difference $H_m - H_1 = (m + 1)ab \left[ \frac{1}{b + ma} - \frac{1}{a + mb} \right]$.$H_m - H_1 = (m + 1)ab \left[ \frac{a + mb - b - ma}{(b + ma)(a + mb)} \right] = (m + 1)ab \frac{(m - 1)(b - a)}{(b + ma)(a + mb)}$.$H_m - H_1 = \frac{(m^2 - 1)ab(b - a)}{m(a^2 + b^2) + (m^2 + 1)ab}$.Using equation $(i)$,the denominator is $m^2 - 1$.Thus,$H_m - H_1 = \frac{(m^2 - 1)ab(b - a)}{m^2 - 1} = ab(b - a)$.

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

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