$\begin{array}{1 1} \large\frac{2n-1}{2n} \\ \large\frac{2n+1}{2n-1} \\ \large\frac{2n-1}{2n+1} \\\large\frac{2n}{2n+1}\end{array} $

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For distance $S_n$ travelled from $(n-1)$ to n,

$S_n= 0 [n-n-1)]+\large\frac{1}{2} $$a[n^2-(n-1)^2]$

$S_n = \large\frac{1}{2}$$a [n^2-(n^2+1-2n)]$

$S_n= \large\frac{a}{2}$$(2n-1)$

Again , $ S_{n+1}= \large\frac{1}{2}$$a[(n+1)^2-n^2]$

$\qquad= \large\frac{a}{2} $$[2n+1]$

$\large\frac{S_n}{S_{n+1}}=\large\frac{\frac {a}{2} (2n-1)}{\frac{a}{2} (2n+1)}$

$\qquad= \large\frac{2n-1}{2n+1}$

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