We are given that $a>b$ and $c<0$.
Multiplying both sides of $a>b$ by a negative number $c$ reverses the inequality.
This is a key rule in inequalities involving negative numbers.
So, $a.c<b.c$ is not true.
But option (C) says $a c>b c$, which is incorrect. Wait – let's recheck.
Since $c<0$ and $a>b$, then multiplying by $c$ gives $a c<b c$.
So the correct relation is $a c<b c$, not $a c>b c$.
Hence, option (C) is actually incorrect.
Let’s now test each option with numbers:
Let $a = 5$, $b = 3$, $c = -2$.
(A): $5 + (-2) = 3$, $3 + (-2) = 1$ → $3<1$ is false.
(B): $5 - (-2) = 7$, $3 - (-2) = 5$ → $7<5$ is false.
(C): $5 \times (-2) = -10$, $3 \times (-2) = -6$ → $-10>-6$ is false.
(D): $5 - (-2) = 7$, $3 + (-2) = 1$ → $7>1$ is true.
If $a>b$ and $c<0$, then $-c$ is positive.
So $a - c$ increases $a$, and $b + c$ reduces $b$.
Therefore, $a - c>b + c$ is a true statement.