Question

If $a_1,a_2,a_3,\dots ,a_n$ are in $A.P$. and $a_1=0$, then the value of $\left(\frac{a_3}{a_2}+\frac{a_4}{a_3}+\cdots +\frac{a_n}{a_{n-1}}\right)-a_2\left(\frac{1}{a_2}+\frac{1}{a_3}+\cdots +\frac{1}{a_{n-2}}\right)$ is equal to

• A. $\left(n-2\right)+\frac{1}{\left(n-2\right)}$
• B. $\frac{1}{n-2}$
• C. $(n-2)$
• D. $n-1$
• E. $n+2$

Answer 1

Answer: (A)

Given; $a_1,a_2,a_3,\dots ,a_n$ are in AP and $a_1=0$
Then, $a_2=a_1+d=0+d=d$
$a_3=a_1+2d=0+2d=2d$
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$a_n=a_1+(n-1)d$
$=(n-1)d$
Now, $\left(\frac{a_3}{a_2}+\frac{a_4}{a_3}+\dots +\frac{a_n}{a_{n-1}}\right)$
$-a_2\left(\frac{1}{a_2}+\frac{1}{a_3}+\dots +\frac{1}{a_{n-2}}\right)$
$=\left(\frac{2d}{d}+\frac{3d}{2d}+\dots +\frac{(n-1)d}{(n-2)d}\right)$
$-d\left(\frac{1}{d}+\frac{1}{2d}+\dots +\frac{1}{(n-3)d}\right)$
$=\left(\frac{2}{1}+\frac{3}{2}+\dots +\frac{(n-1)}{(n-2)}\right)$
$-\left(1+\frac{1}{2}+\dots +\frac{1}{n-3}\right)$
$$\begin{gathered} =\left\{(1+1)+\left(1+\frac{1}{2}\right)+\dots +\left(1+\frac{1}{n-2}\right)\right\} \\ -\left\{1+\frac{1}{2}+\dots +\frac{1}{n-3}\right\} \end{gathered}$$
$=\{(1+1+1+\dots +(n-2)$ terms )
$\begin{array}{rl}+(1+\frac{1}{2}+\frac{1}{3}& +\dots +\frac{1}{n-3})+\frac{1}{n-2}\}\\ -\left\{1+\frac{1}{2}+\dots +\frac{1}{n-3}\right\}& \end{array}\}$
$=\{(1+1+\dots +(n-2)$ terms $)+\frac{1}{n-2}\}$
$=(n-2)+\frac{1}{(n-2)}$