1. Represent the five terms of the A.P.

For five numbers in A.P., let them be $a-2d,\ a-d,\ a,\ a+d,\ a+2d.$

2. Use the sum condition

Their sum is $25$, so $(a-2d)+(a-d)+a+(a+d)+(a+2d)=5a=25$ Hence, $a=5.$

So the five numbers are $5-2d,\ 5-d,\ 5,\ 5+d,\ 5+2d.$

3. Use the condition that one term is $-\frac12$

Since one of the five numbers equals $-\frac12$, we check possible positions:

• $5=-\frac12$ is impossible.
• If $5-d=-\frac12$, then $d=\frac{11}{2}$.
• If $5-2d=-\frac12$, then $2d=\frac{11}{2}$, so $d=\frac{11}{4}$.
• If $5+d=-\frac12$ or $5+2d=-\frac12$, then $d<0$, which just reverses the same A.P. already covered.

So we only need to test $d=\frac{11}{2}$ and $d=\frac{11}{4}$ using the product condition.

4. Check the product

The product is given as $2520$.

Case 1: $d=\frac{11}{2}$

Then the numbers are $5-11=-6,\quad 5-\frac{11}{2}=-\frac12,\quad 5,\quad 5+\frac{11}{2}=\frac{21}{2},\quad 5+11=16.$ Their product is $(-6)\left(-\frac12\right)(5)\left(\frac{21}{2}\right)(16).$ Now, $(-6)\left(-\frac12\right)=3,$ so $3\cdot 5\cdot \frac{21}{2}\cdot 16=15\cdot 168=2520.$ This satisfies the condition.

Case 2: $d=\frac{11}{4}$

Then the numbers are $5-\frac{11}{2}=-\frac12,\quad 5-\frac{11}{4}=\frac94,\quad 5,\quad \frac{31}{4},\quad \frac{21}{2}.$ Their product is $-\frac12\cdot \frac94\cdot 5\cdot \frac{31}{4}\cdot \frac{21}{2},$ which is negative, so it cannot be $2520$.

Thus only Case 1 is valid.

5. Find the greatest number

The valid A.P. is $-6,\ -\frac12,\ 5,\ \frac{21}{2},\ 16.$ Hence the greatest number is $\boxed{16}.$

6. Compare with stored answer

Stored correct answer: D

Our derived answer is D: $16$, so it agrees.