Explanation

(b)
We have, $\frac{1}{x_{1}},\frac{1}{x_{2}},\frac{1}{x_{3}},\ldots,\frac{1}{x_{n}}$ are in AP.
$\therefore\frac{1}{x_{2}} - \frac{1}{x_{1}} = \frac{1}{x_{3}} - \frac{1}{x_{2}} = ... = \frac{1}{x_{n}} - \frac{1}{x_{n - 1}} = d$ (say)
$\therefore\ \frac{x_{1} - x_{2}}{x_{1}x_{2}} = \frac{x_{2} - x_{3}}{x_{2}x_{3}} = ... = \frac{x_{m - 1} - x_{n}}{x_{n - 1}x_{n}} = d$
Now, $x_{1}x_{2} + x_{2}x_{3} + ... + x_{n + 1}x_{n}$
$= \frac{1}{d}\left\lbrack x_{1} - x_{2} + x_{2} - x_{3} + \ldots + x_{n - 1} - x_{n} \right\rbrack$
$= \ \frac{x_{1} - x_{n}}{d}$
But $\frac{1}{x_{n}} = \frac{1}{x_{1}} + (n - 1)d$
$\therefore\ \ \frac{x_{1} - x_{n}}{x_{1}x_{n}} = (n - 1)d$
or $\frac{x_{1} - x_{n}}{d} = (n - 1)x_{1}x_{n}$
$\therefore x_{1}x_{2} + x_{2}x_{3} + ... + x_{n - 1}x_{n} = (n - 1)x_{1}x_{n}$