Sum of Squares of Terms of Arithmetic Progression

Let the sum of the first $n$ terms of the Arithmetic Progression (A.P.) be denoted by $S_n$.
Given:
$S_n = cn^2$

The $r^{\text{th}}$ term $T_r$ of this A.P. can be determined using the relation $T_r = S_r - S_{r-1}$ for $r \ge 1$:
$T_r = c r^2 - c(r-1)^2 = c\left[r^2 - (r^2 - 2r + 1)\right] = c(2r - 1)$

We need to find the sum of the squares of the first $n$ terms, denoted by $\Sigma$:
$\Sigma = \sum_{r=1}^{n} T_r^2 = \sum_{r=1}^{n} \left[c(2r - 1)\right]^2 = c^2 \sum_{r=1}^{n} (2r - 1)^2$

Expanding the term inside the summation:
$(2r - 1)^2 = 4r^2 - 4r + 1$

Thus,
$\sum_{r=1}^{n} (2r - 1)^2 = 4\sum_{r=1}^{n} r^2 - 4\sum_{r=1}^{n} r + \sum_{r=1}^{n} 1$

Using standard summation formulas:
$\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$
$\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$
$\sum_{r=1}^{n} 1 = n$

Substituting these expressions:
$\sum_{r=1}^{n} (2r - 1)^2 = 4 \left[\frac{n(n+1)(2n+1)}{6}\right] - 4 \left[\frac{n(n+1)}{2}\right] + n$
$= \frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n$
$= \frac{n}{3} \left[2(n+1)(2n+1) - 6(n+1) + 3\right]$
$= \frac{n}{3} \left[2(2n^2 + 3n + 1) - 6n - 6 + 3\right]$
$= \frac{n}{3} \left[4n^2 + 6n + 2 - 6n - 3\right]$
$= \frac{n(4n^2 - 1)}{3}$

Therefore, the sum of the squares of these $n$ terms is:
$\Sigma = \frac{n(4n^2 - 1)c^2}{3}$

Hence, the correct option is (C).