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nth Term of A.P

If a1,a2,a3...

Question

If $a_{1}, a_{2}, a_{3}, . . . . . . a_{n}$ are in H.P then $a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} . . . . + a_{n - 1} a_{n} =$ ?

A

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

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B

(n + 1) a_{1} a_{n}

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C

(n - 2) a_{1} a_{n}

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D

(n + 4) a_{1} a_{n}

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Solution

The correct option is C $(n - 1) a_{1} a_{n}$
Since $a_{1}, a_{2}, a_{3}, . . . a_{n}$ are in H.P
$∴ \frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}}, . . ., \frac{1}{a_{n}}$ are in A.P
Let $d$ be the common difference of AP
$∴ \frac{1}{a_{2}} - \frac{1}{a_{1}} = d$
$\Rightarrow \frac{a_{1} - a_{2}}{a_{1} a_{2}} = d$
$\Rightarrow a_{1} - a_{2} = a_{1} a_{2} d$
$\Rightarrow a_{2} - a_{3} = a_{2} a_{3} d$
..............................................
$a_{n - 1} - a_{n} = a_{n - 1} a_{n} d$
On adding all of these , we get
$a_{1} - a_{n} = d (a_{1} a_{2} + a_{2} a_{3} + . . . + a_{n - 1} a_{n})$ ....... $(1)$
Also, $\frac{1}{a_{n}} = \frac{1}{a_{1}} + (n - 1) d$
$\Rightarrow d = \frac{a_{1} - a_{n}}{a_{1} a_{n} (n - 1)}$
On putting the value of d in Eq, $(i)$ , we get
$a_{1} - a_{n} = \frac{a_{1} - a_{n}}{a_{1} a_{n} (n - 1)} (a_{1} a_{2} + a_{2} a_{3} + . . . + a_{n - 1} a_{n})$
$\Rightarrow a_{1} a_{2} + a_{2} a_{3} + . . . + a_{n - 1} a_{n} = a_{1} a_{n} (n - 1)$

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