If the sum of n terms of an A.P. is an(n - 1), then sum of squares of these n terms is

Options

A. `(2a^(2))/(3) n (n - 1) (2n - 1)`
B. `(2a^(2))/(3) n(n + 1) (2n + 1)`
C. `(a^(2))/(6) n (n - 1) (2n - 1)`
D. `(a^(2))/(6) n (n + 1) (2n + 1)`

Answer

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

### Step-by-Step Solution:

1. **Understanding the Sum of n Terms of A.P.**:
The sum of the first \( n \) terms of an A.P. can be expressed as:
\[ S_n = \frac{n}{2} \left(2a + (n - 1)d\right) \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Setting the Given Condition**:
According to the problem, we have:
\[ S_n = a n (n - 1) \]
We can equate the two expressions for \( S_n \):
\[ \frac{n}{2} \left(2a + (n - 1)d\right) = a n (n - 1) \]

3. **Simplifying the Equation**:
Multiply both sides by 2 to eliminate the fraction:
\[ n(2a + (n - 1)d) = 2a n (n - 1) \]
Dividing both sides by \( n \) (assuming \( n \neq 0 \)):
\[ 2a + (n - 1)d = 2a(n - 1) \]

4. **Rearranging the Equation**:
Rearranging gives:
\[ (n - 1)d = 2a(n - 1) - 2a \]
Simplifying further:
\[ (n - 1)d = 2a(n - 2) \]
Thus, we can express \( d \):
\[ d = \frac{2a(n - 2)}{n - 1} \]

5. **Finding the n-th Term**:
The \( n \)-th term \( T_n \) of the A.P. is given by:
\[ T_n = a + (n - 1)d \]
Substituting the value of \( d \):
\[ T_n = a + (n - 1) \left(\frac{2a(n - 2)}{n - 1}\right) \]
Simplifying gives:
\[ T_n = a + 2a(n - 2) = 2an - 3a \]

6. **Sum of Squares of n Terms**:
The sum of squares of the first \( n \) terms can be expressed as:
\[ \sum_{k=0}^{n-1} T_k^2 \]
Using the formula for the sum of squares:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]
We need to calculate:
\[ \sum_{k=0}^{n-1} (2ak - a)^2 = \sum_{k=0}^{n-1} (2ak - a)^2 = 4a^2 \sum_{k=0}^{n-1} k^2 - 4a^2 \sum_{k=0}^{n-1} k + na^2 \]

7. **Final Calculation**:
Plugging in the values for the sums:
\[ S = 4a^2 \left(\frac{(n-1)n(2(n-1)+1)}{6}\right) - 4a^2 \left(\frac{(n-1)n}{2}\right) + na^2 \]

8. **Conclusion**:
After simplifying, we find that the sum of the squares of the first \( n \) terms of the A.P. is:
\[ S = \frac{n(n-1)(2n-1)}{6} \cdot 4a^2 \]

### Final Answer:
The sum of the squares of the first \( n \) terms of the A.P. is:
\[ \frac{2}{3} a^2 n(n-1)(2n-1) \]