The number of terms in an A.P. (Arithmetic Progression) is even. The sum of the odd and even-numbered terms are 24 and 30, respectively. If the last term exceeds the first by 10.5, the number of terms in the A.P. is 
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 8$
Let $a$ be the first term, $n$ be the number of terms, and $d$ be the common difference.  That means the last term is $a+r(n-1)$.
We can write an equation on the difference between the last and first term based on the conditions.
\[a+r(n-1)-a =10.5\]
\[rn-r=10.5\]
Also, half of the terms add up to $24$ while the other half of the terms add up to $30$, so
\[24 + r\frac{n}{2} = 30\]
\[nr = 12\]
Substituting the value back to a previous equation,
\[12-r=10.5\]
\[r=1.5\]
Substituting to a previous equation again,
\[1.5n-1.5=10.5\]
\[n=8\]
Thus, there are $\boxed{\textbf{(E)}\ 8}$ terms in the arithmetic sequence.