Question

If $a_1,a_2,............a_n$ are in H.P., then the expression $a_1{a}_2+a_2{a}_3+............+a_{n-1}{a}_n$ is equal to

• A. $(n-1)a_1{a}_n$
• B. $n(a_1-a_n)$
• C. $(n-1)(a_1-a_n)$
• D. $n{a}_1{a}_n$

Answer 1

Answer: (A)

Since $a_1,a_2,a_3,......a_n$ are in $H.P$.
$\therefore \frac{1}{a_1},\frac{1}{a_2},....,\frac{1}{a_n}$ are in $A.P$.
Let $R$ be common difference of the $A.P$.
Now $a_1{a}_2+a_2{a}_3+.....+a_{n-1}{a}_n$
$=\frac{1}{d}\left[a_1-a_2+a_2-a_3+.....a_{n-1}-a_n\right]$
$[\frac{1}{a_2}-\frac{1}{a_1}=d_1\Rightarrow \frac{a_1-a_2}{a_2{a}_1}=d\frac{a_1-a_2}{d}=a_1{a}_2$ etc.]etc.
$=\frac{1}{d}\left(a_1-a_n\right)=\frac{1}{d}\left[\frac{a_1-a_n}{\left(n-1\right)}\left(n-1\right)\right]$
$=\frac{\left(n-1\right)\left(a_1-a_n\right)}{\frac{1}{a_n}-\frac{1}{a_1}}\left[\because \frac{1}{a_n}=\frac{1}{a_1}+\left(n-1\right)d\right]$
$=\left(n-1\right)a_1{a}_n$