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

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