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  4. If a1, a2, a3, ldots, an are in A.P. and a1=0, then the value of ((a3/a2)+(a4/a3)+⋅s+(an/an-1))-a2 ((1/a2)+(1/a3)+⋅s+(1/an-2)) is equal to

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Q. If $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ 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

KEAMKEAM 2011Sequences and Series

Solution:

Given; $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ are in AP and $a_{1}=0$
Then, $a_{2}=a_{1}+d=0+d=d$
$a_{3}=a_{1}+2 d=0+2 d=2 d$
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$ a_{n} =a_{1}+(n-1) d $
$=(n-1) d $
Now, $\left(\frac{a_{3}}{a_{2}}+\frac{a_{4}}{a_{3}}+\ldots+\frac{a_{n}}{a_{n-1}}\right)$
$-a_{2}\left(\frac{1}{a_{2}}+\frac{1}{a_{3}}+\ldots+\frac{1}{a_{n-2}}\right)$
$=\left(\frac{2 d}{d}+\frac{3 d}{2 d}+\ldots+\frac{(n-1) d}{(n-2) d}\right)$
$-d\left(\frac{1}{d}+\frac{1}{2 d}+\ldots+\frac{1}{(n-3) d}\right)$
$=\left(\frac{2}{1}+\frac{3}{2}+\ldots+\frac{(n-1)}{(n-2)}\right) $
$-\left(1+\frac{1}{2}+\ldots+\frac{1}{n-3}\right) $
$=\left\{(1+1)+\left(1+\frac{1}{2}\right)+\ldots+\left(1+\frac{1}{n-2}\right)\right\} \\-\left\{1+\frac{1}{2}+\ldots+\frac{1}{n-3}\right\}$
$=\{(1+1+1+\ldots+(n-2)$ terms )
$\left.\begin{array}{rl}+\left(1+\frac{1}{2}+\frac{1}{3}\right. & \left. \left.+\ldots+\frac{1}{n-3}\right)+\frac{1}{n-2}\right\} \\ -\left\{1+\frac{1}{2}+\ldots+\frac{1}{n-3}\right\}\end{array}\right\}$
$=\left\{(1+1+\ldots+(n-2)\right.$ terms $\left.)+\frac{1}{n-2}\right\}$
$=(n-2)+\frac{1}{(n-2)}$