Indian Institute of Foreign Trade 2008 Solved Paper

Let us assume that n = 3 and a, b = 2, 6.
Therefore, the harmonic sequence will be: 2, ${\mathit{H}}_1$, ${\mathit{H}}_2$, ${\mathit{H}}_3$,6
Hence, ${\mathit{H}}_2$ = $\frac{\left(2*2*6\right)}{\left(2+6\right)}$ =3
${\mathit{H}}_1$ = $\frac{\left(2*2*3\right)}{\left(2+3\right)}$ =2.4
${\mathit{H}}_3$ = $\frac{\left(2*3*6\right)}{\left(3+6\right)}$ =4
Therefore, $\frac{\left({\mathit{H}}_1+\mathit{a}\right)}{\left({\mathit{H}}_1-\mathit{a}\right)}$+$\frac{\left({\mathit{H}}_{\mathit{n}}+\mathit{b}\right)}{\left({\mathit{H}}_{\mathit{n}}-\mathit{b}\right)}$
$\frac{\left(2.4+2\right)}{\left(2.4-2\right)}$+$\frac{\left(4+6\right)}{\left(4-6\right)}$
⇒11-5 = 6
Option B: 2n = 2*3 = 6.