If $a_{1}, a_{2}, a_{3}, . . . . a_{n}$ are in HP, then $a_{1} a_{2} + a_{2} a_{3} + . . . . . + a_{n - 1} a_{n}$ will be equal to

a_{1} a_{n}

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n a_{1} a_{n}

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(n - 1) a_{1} a_{n}

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None of these

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Solution

The correct option is C $(n - 1) a_{1} a_{n}$
Given $a_{1}, a_{2}, a_{3}, . . . . . . a_{n}$ are in HP
Then, $\frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}}, . . . . . \frac{1}{a_{n}}$ will be in AP
which gives
$\frac{1}{a_{2}} - \frac{1}{a_{1}} = \frac{1}{a_{3}} - \frac{1}{a_{2}} = . . . . . = \frac{1}{a_{n}} - \frac{1}{a_{n - 1}} = d$
$\Rightarrow \frac{a_{1} - a_{2}}{a_{1} a_{2}} = \frac{a_{2} - a_{3}}{a_{2} a_{3}} = . . . . . . = \frac{a_{n - 1} - a_{n}}{a_{n - 1} a_{n}} = d$
$\Rightarrow a_{1} - a_{2} = d a_{1} a_{2}$
$a_{2} - a_{3} = d a_{2} a_{3}$
........
....
$a_{n - 1} - a_{n} = d a_{n - 1} a_{n}$
On adding these, we have
$d (a_{1} a_{2} + a_{2} a_{3} + . . . . + a_{n - 1} a_{n}) = a_{1} - a_{n}$ ....(i)
Also $n^{t h}$ term of this AP is given by
$\frac{1}{a_{n}} = \frac{1}{a_{1}} + (n - 1) d$
$d = \frac{a_{1} - a_{n}}{a_{1} a_{n} (n - 1)}$
On substituting this value of $d$ in Eq. (i), we get
$(a_{1} - a_{n}) = \frac{a_{1} - a_{n}}{a_{1} a_{n} (n - 1)} (a_{1} a_{2} + a_{2} a_{3} + . . . . + a_{n - 1} a_{n})$
$\Rightarrow (a_{1} a_{2} + a_{2} a_{3} + . . . . + a_{n - 1} a_{n}) = a_{1} a_{n} (n - 1)$

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