Question

If $a_{1}, a_{2}, \ldots, a_{n} $ are in HP, then the expression $a_{1} a_{2}+a_{2} a_{3}+.......+a_{n-1} a_{n} $ is equal to :

• A. $\left(n-1\right) \left(a_{1}-a_{n}\right)$
• B. $n a_{1}a_{n}$
• C. $\left(n-1\right)a_{1} a_{n}$
• D. $n \left(a_{1}-a_{n}\right)$

Answer 1

Answer: (C)

$\because 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 the common difference of AP.
$\therefore \, \frac{1}{a_{2}}-\frac{1}{a_{1}}=d \Rightarrow a_{1}-a_{2}=a_{1}a_{2}d$
Similarly $\, $ $a_{2}-a_{3}=a_{2}a_{3} d$
$\ldots\ldots\ldots\ldots\ldots$
$\ldots\ldots\ldots\ldots\ldots$
$a_{n-1} -a_{n}=a_{n-1} a_{n} d$
On adding all of these, we get
$a_{1}-a_{n}=d \left\{a_{1}a_{2}+a_{2}a_{3}+\ldots\ldots+a_{n-1} a_{n}\right\} \ldots$ (i)
Also $\frac{1}{a_{n}}=\frac{1}{a_{1}}+\left(n-1\right)d$
$\Rightarrow \, d=\frac{a_{1-a_n}}{a_{1} a_{n} \left(n-1\right)}$
On putting this value of d in equation (i), we get
$a_{1}-a_{n}=\frac{a_{1}-a_{n}}{a_{1}a_{n}\left(n-1\right)} \left\{a_{1}a_{2}+a_{2}a_{3}+\ldots\ldots +a_{n-1} a_{n} \right\}$
$\Rightarrow \, a_{1}a_{2}+a_{2}a_{3}+\ldots+a_{n-1} a_{n}=a_{1}a_{n} \left(n-1\right)$