If m is a root of the given equation (1 - ab) | vedclass

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(B) Given the equation $(1 - ab)m^2 - (a^2 + b^2)m - (1 + ab) = 0$.Rearranging terms,we get $m(a^2 + b^2) + (m^2 + 1)ab = m^2 - 1$ ......$(i)$.Let $H_

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$ab(b - a)$

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(B) Given the equation $(1 - ab)m^2 - (a^2 + b^2)m - (1 + ab) = 0$.Rearranging terms,we get $m(a^2 + b^2) + (m^2 + 1)ab = m^2 - 1$ ......$(i)$.Let $H_1, H_2, \dots, H_m$ be the $m$ harmonic means between $a$ and $b$. The sequence $a, H_1, H_2, \dots, H_m, b$ is in Harmonic Progression ($H$.$P$.).Thus,$\frac{1}{a}, \frac{1}{H_1}, \dots, \frac{1}{H_m}, \frac{1}{b}$ is in Arithmetic Progression ($A$.$P$.).Let $d$ be the common difference of this $A$.$P$. Then $\frac{1}{b} = \frac{1}{a} + (m + 1)d$,so $d = \frac{a - b}{ab(m + 1)}$.$H_1 = \frac{1}{\frac{1}{a} + d} = \frac{ab(m + 1)}{b(m + 1) + a - b} = \frac{ab(m + 1)}{mb + a}$.$H_m = \frac{1}{\frac{1}{b} - d} = \frac{ab(m + 1)}{a(m + 1) - (a - b)} = \frac{ab(m + 1)}{ma + b}$.Difference $H_m - H_1 = ab(m + 1) \left[ \frac{1}{ma + b} - \frac{1}{mb + a} \right] = ab(m + 1) \frac{mb + a - ma - b}{(ma + b)(mb + a)} = ab(m + 1) \frac{(m - 1)(b - a)}{m^2ab + ma^2 + mb^2 + ab} = \frac{ab(m^2 - 1)(b - a)}{m(a^2 + b^2) + ab(m^2 + 1)}$.Using equation $(i)$,the denominator is $m^2 - 1$.Thus,$H_m - H_1 = \frac{ab(m^2 - 1)(b - a)}{m^2 - 1} = ab(b - a)$.

If $m$ is a root of the given 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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