Let $a$ and $d$ be the first term and the common difference of the arithmetic sequence respectively.
I may not be true. If the $d< 0$, then $x_1 > x_2$.
II must be true. The range of the first group
$\begin{array}{cl}
= & T(49) – T(1) \\
= & a +48d -a \\
= & 48d
\end{array}$
The range of the second group
$\begin{array}{cl}
= & T(99) -T(51) \\
= & a +98d -a -50d \\
= & 48d
\end{array}$
Hence, $y_1 = y_2$.
III is false. It is the fact that adding a constant to each datum of a group of number does not affect the value of the range, the inter-quartile range, the standard deviation and the variance.
Note that
$\begin{array}{rcl}
T(51) & = & T(1) +50d \\
T(52) & = & T(2) +50d \\
& \vdots & \\
T(99) & = & T(49) +50d
\end{array}$
Therefore, the second group is created by adding $50d$ to each datum of the first group. Hence, $z_1 =z_2$.
