Given that sum from r=1 to 30 of (5a + 4r) = 3960, where a is a constant. Find the value of a. Using the value of a found in (a), find an expression in terms of n for sum from r=5 to n of (5a + 4r). Sketch the graphs of y = Sin x and y = pi/2 - 2x for 0 <= x <= pi/2 on the same axis.

The given equation is $\sum_{r=1}^{30} (5a + 4r) = 3960$. We can split the summation into two parts:

For the first part, $5a$ is a constant with respect to $r$, so:

For the second part, we use the formula for the sum of the first $N$ natural numbers, $\sum_{r=1}^{N} r = \frac{N(N+1)}{2}$:

The value of $a$ is $\boxed{14}$.

Using $a=14$ from part (a), the expression becomes:

This is an arithmetic series with the first term (when $r=5$) $T_5 = 70 + 4(5) = 70 + 20 = 90$.
The last term (when $r=n$) is $T_n = 70 + 4n$.
The number of terms in the series is $k = n - 5 + 1 = n - 4$.

The sum of an arithmetic series is given by $S_k = \frac{k}{2}(\text{first term} + \text{last term})$.

Factor out 2 from $(160 + 4n)$:

Expand the product:

Combine like terms:

The expression in terms of $n$ is $\boxed{2n^2 + 72n - 320}$.