If the sum of n terms of an A.P is cn (n-1)where `c ne 0` then the sum of the squares of these terms is

Options

A. `c^2n(n+1)^2`
B. `2/3c^2n(n-1)(2n-1)`
C. `(2c^2)/(3)n(n+1)(2n+1)`
D. none of these

Explanation

To solve the problem, we need to find the sum of the squares of the terms of an arithmetic progression (A.P.) given that the sum of the first n terms is \( S_n = cn(n-1) \), where \( c \neq 0 \).

### Step-by-step Solution:

1. **Identify the Sum of n Terms**:
The sum of the first n terms of the A.P. is given as:
\[ S_n = cn(n-1) \]

2. **Find the nth Term**:
The nth term \( t_n \) can be found using the formula:
\[ t_n = S_n - S_{n-1} \]
We first need to find \( S_{n-1} \):
\[ S_{n-1} = c(n-1)((n-1)-1) = c(n-1)(n-2) \]
Now substituting back:
\[ t_n = S_n - S_{n-1} = cn(n-1) - c(n-1)(n-2) \]
Simplifying this:
\[ t_n = c(n-1) \left[ n - (n-2) \right] = c(n-1)(2) = 2c(n-1) \]

3. **Sum of the Squares of the Terms**:
We need to find the sum of the squares of the first n terms:
\[ \text{Sum of squares} = t_1^2 + t_2^2 + t_3^2 + \ldots + t_n^2 \]
We can express \( t_k \) for \( k = 1, 2, \ldots, n \):
\[ t_k = 2c(k-1) \]
Therefore:
\[ t_k^2 = (2c(k-1))^2 = 4c^2(k-1)^2 \]
Thus, the sum of squares becomes:
\[ \sum_{k=1}^{n} t_k^2 = \sum_{k=1}^{n} 4c^2(k-1)^2 = 4c^2 \sum_{k=1}^{n} (k-1)^2 \]

4. **Using the Formula for the Sum of Squares**:
The sum of the squares of the first \( n-1 \) natural numbers is given by:
\[ \sum_{j=0}^{n-1} j^2 = \frac{(n-1)n(2(n-1)+1)}{6} \]
Therefore:
\[ \sum_{k=1}^{n} (k-1)^2 = \sum_{j=0}^{n-1} j^2 = \frac{(n-1)n(2(n-1)+1)}{6} \]

5. **Final Expression for the Sum of Squares**:
Substituting back into the sum of squares:
\[ \sum_{k=1}^{n} t_k^2 = 4c^2 \cdot \frac{(n-1)n(2(n-1)+1)}{6} \]
Simplifying this gives:
\[ = \frac{2}{3}c^2(n-1)n(2n-1) \]

### Final Result:
The sum of the squares of the terms is:
\[ \frac{2}{3}c^2(n-1)n(2n-1) \]