NIMCET 2008 Question Paper with solutions

a. ${\mathit{H}}_1,{\mathit{H}}_2,...,{\mathit{H}}_{\mathit{n}},\mathit{b}$ in HP.
$\Rightarrow \frac{1}{\mathit{a}},\frac{1}{{\mathit{H}}_1},\frac{1}{{\mathit{H}}_2}...\frac{1}{{\mathit{H}}_{\mathit{n}}}\frac{1}{\mathit{b}}$ in AP.
$\Rightarrow \frac{1}{\mathit{b}}=\frac{1}{\mathit{a}}+\left(\mathit{n}+2-1\right)\mathit{d}$
$\Rightarrow \frac{1}{\mathit{b}}-\frac{1}{\mathit{a}}=\left(\mathit{n}+1\right)\mathit{d}$
$\Rightarrow \mathit{d}=\frac{\frac{1}{\mathit{b}}-\frac{1}{\mathit{a}}}{\left(\mathit{n}+1\right)}$
$\therefore \frac{1}{{\mathit{H}}_1}=\frac{1}{\mathit{a}}+\mathit{d}=\frac{1}{\mathit{a}}+\frac{\frac{1}{\mathit{b}}-\frac{1}{\mathit{a}}}{\mathit{n}+1}$
$=\frac{1}{\mathit{a}}+\frac{\mathit{a}-\mathit{b}}{\mathit{a}\mathit{b}\left(\mathit{n}+1\right)}$
$=\frac{\mathit{b}\left(\mathit{n}+1\right)+\mathit{a}-\mathit{b}}{\mathit{a}\mathit{b}\left(\mathit{n}+1\right)}$
$\Rightarrow {\mathit{H}}_1=\frac{\mathit{a}\mathit{b}\left(\mathit{n}+1\right)}{\mathit{b}\mathit{n}+\mathit{a}}$
Using componendo and Dividendo, we get
$\Rightarrow \frac{{\mathit{H}}_{\mathit{n}}+\mathit{b}}{{\mathit{H}}_{\mathit{n}}-\mathit{b}}=\frac{\mathit{a}\mathit{n}+\mathit{a}+\mathit{a}\mathit{n}+\mathit{b}}{\mathit{a}\mathit{n}+\mathit{a}-\mathit{a}\mathit{n}-\mathit{b}}$
$=\frac{\mathit{a}+\mathit{b}+2\mathit{b}\mathit{n}}{\mathit{b}-\mathit{a}}$ .....(i)
Also, $\frac{1}{{\mathit{H}}_{\mathit{n}}}=\frac{1}{\mathit{b}}-\mathit{d}=\frac{1}{\mathit{b}}-\frac{\mathit{a}-\mathit{b}}{\mathit{a}\mathit{b}\left(\mathit{n}+1\right)}$
$=\frac{\mathit{a}\left(\mathit{n}+1\right)-\mathit{a}+\mathit{b}}{\mathit{a}\mathit{b}\left(\mathit{n}+1\right)}$
$\Rightarrow {\mathit{H}}_{\mathit{n}}=\frac{\mathit{a}\mathit{b}\left(\mathit{n}+1\right)}{\mathit{a}\mathit{n}+\mathit{b}}$
$\Rightarrow \frac{\mathit{H}\mathit{n}}{\mathit{b}}=\frac{\mathit{a}\mathit{n}+\mathit{a}}{\mathit{a}\mathit{n}+\mathit{b}}$
Using componendo and Dividendo, we get
$\Rightarrow \frac{{\mathit{H}}_{\mathit{n}}+\mathit{b}}{{\mathit{H}}_{\mathit{n}}-\mathit{b}}=\frac{\mathit{a}\mathit{n}+\mathit{a}+\mathit{a}\mathit{n}+\mathit{b}}{\mathit{a}\mathit{n}+\mathit{a}-\mathit{a}\mathit{n}-\mathit{b}}$
$=\frac{\mathit{a}+\mathit{b}+2\mathit{a}\mathit{n}}{\mathit{a}-\mathit{b}}$ .....(ii)
On adding Eqs. (i) and (ii),
$=\frac{2\mathit{n}\left(\mathit{b}-\mathit{a}\right)}{\mathit{b}-\mathit{a}}=2\mathit{n}$