Question

If $a_1,a_2,\dots ,a_n$ 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)

$\because a_1,a_2,a_3,\dots ,a_n$ are in HP
$\therefore \frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3},\dots ,\frac{1}{a_n}$ 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 $a_2-a_3=a_2{a}_{3}d$
$\dots \dots \dots \dots \dots$
$\dots \dots \dots \dots \dots$
$a_{n-1}-a_n=a_{n-1}{a}_{n}d$
On adding all of these, we get
$a_1-a_n=d\left\{a_1{a}_2+a_2{a}_3+\dots \dots +a_{n-1}{a}_n\right\}\dots$ (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
$a_1-a_n=\frac{a_1-a_n}{a_1{a}_n\left(n-1\right)}\left\{a_1{a}_2+a_2{a}_3+\dots \dots +a_{n-1}{a}_n\right\}$
$\Rightarrow a_1{a}_2+a_2{a}_3+\dots +a_{n-1}{a}_n=a_1{a}_n\left(n-1\right)$