Can we express median in terms of standard operations?

Yes! Write $n=2k+1$. Then the median is the maximum of the minima of all $k+1$ element subsets of the data.

For example, when $n=3$, then there are $\binom{3}2=3$ two element subsets of the data: $\{a_1,a_2\},\{a_1,a_3\},\{a_2,a_3\}$. The median is the max of all these minima: $$ \text{median}(a_1,a_2,a_3)=\max\big(\min(a_1,a_2),\min(a_1,a_3),\min(a_2,a_3)\big) $$ When $n=5$, there are $\binom53=10$ three element subsets of the data, so $$ \text{median}=\max\Big(\min(a_1,a_2,a_3),\min(a_1,a_2,a_4),\min(a_1,a_2,a_5),\min(a_2,a_3,a_4),\dots,\min(a_3,a_4,a_5)\Big) $$ As you can see, this formula quickly gets unwieldy. The number of terms is $\binom{n}{(n+1)/2}$, which grows exponentially.

The median is the arg min of the sum of absolute deviations of $X_i$ from $\theta.$

Similarly, the mean is analogous for squared deviation.

$F(k)=\sum_{r=1}^n |k-a_r| $, (where k is a variabe , $a_r>a_{r+1}$)

If n is odd ,n=2m+1

$k>a_1\Rightarrow$ all mods open as $k-a_r$ and so derivative is >0(as you increase 'k' the value of function increase).

$a_1>k>a_2$ then one mod only open as $a_r-k$(r=1) , and others as $k-a_r$, hence still derivative >0

$a_m>k>a_{m+1} $ ,here m+1 mods open as $k-a_r$ and m mods open as $a_r-k$ ,so derivative is greater than 0

So obviusly derivative becomes negative if $k<a_{m+1}$ and derivative is greater than 0 if $k>a_{m+1}$ , hence minimum of the function is at $a_{m+1}$.

If n is even ,n=2m

You will find that derivative is >0 if , $k>a_{m}$

Derivative <0 if $k<a_{m+1}$

Derivative =0 if $ a_m>k>a_{m+1}$ , so the function is minimum for all such $k$(including $a_m$ and $a_{m+1}$).

So for even number of terms we define mode as $\frac{a_m+a_{m+1}}{2}.$