Explanation

(c)
Since, $a_{1},a_{2},a_{3},\ldots.,a_{n}$ are in HP.
$\therefore\frac{1}{a_{1}},\frac{1}{a_{2}},\frac{1}{a_{3}},\ldots,\frac{1}{a_{n}}$ are in AP.
Let $d$ be common difference of AP, then $\frac{1}{a_{2}} - \frac{1}{a_{1}} = d$
$\Rightarrow \ a_{1} - a_{2} = a_{1}a_{2}d$
Similarly, $a_{2} - a_{3} = a_{2}a_{3}d$
$\ldots\ \ldots.\ \ldots.\ \ldots.$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots\ \ldots.\ \ldots.\ \ldots.$
$a_{n - 1} - a_{n} = a_{n - 1}a_{n}d$
On adding all of these, we get
$a_{1} - a_{n} = d(a_{1}a_{2} + a_{2}a_{3} + ... + a_{n - 1}a_{n})$ …(i)
Also, $\frac{1}{a_{n}} = \frac{1}{a_{1}} + (n - 1)d$
$\Rightarrow d = \frac{a_{1} - a_{2}}{a_{1}a_{n}(n - 1)}$
On putting the value of $d$ in Eq. (i), we get
$a_{1} - a_{n} = \frac{a_{1} - a_{n}}{a_{1}a_{n}(n - 1)}(a_{1}a_{2} + a_{2}a_{3} + ... + a_{n - 1}a_{n})$
$\Rightarrow \ a_{1}a_{2} + a_{2}a_{3} + ... + a_{n - 1}a_{n} = a_{1}a_{n}(n - 1)$