Question

If $a_1,a_2,............ a_n$ are in H.P., then the expression $a_1a_2+a_2a_3 +............+ a_{n-1}a_n$ is equal to

• A. $(n-1)a_1a_n$
• B. $n(a_1-a_n)$
• C. $(n-1)(a_1-a_n)$
• D. $na_1a_n$

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} $