Question

( Rightarrow|z-1|+|z-2|+|z-3|+|z-4|+|z-5| )
( therefore ) As all functions are modulus
( therefore ) we choose middle value ( 6 cdot 3=z )
( therefore(3-i|+| 3-2|+| 3-3)+(3-4)+|3-5| )
( =6 mid quad(4) 6 )

# (20) The minimum value of 12-1+2–2 + 2-3 + 2–4 +2–5 is 1) 3 2) 5 37 146

Solution