Let n quantities be in A.P., d being the common difference. Let the arithmetic mean of the squares of these quantities exceed the square of the arithmetic mean of these quantities by a quantity p. Then p

Options

A. is always negative
B. equals `(n^(2)- 1)/(12) d^(2)`
C. equal s`(d^(2))/(12)`
D. equals `(n^(2) - 1)/(12)`

Answer

To find the value of \( p \) given that the arithmetic mean of the squares of \( n \) quantities in an arithmetic progression (A.P.) exceeds the square of the arithmetic mean of these quantities by \( p \), we can follow these steps:

### Step 1: Define the quantities in A.P.
Let the \( n \) quantities in A.P. be represented as:
\[ a, a + d, a + 2d, \ldots, a + (n-1)d \]

### Step 2: Calculate the arithmetic mean (AM)
The arithmetic mean \( A_m \) of these quantities is given by:
\[ A_m = \frac{a + (a + d) + (a + 2d) + \ldots + (a + (n-1)d)}{n} \]
The sum of the terms can be simplified:
\[ = \frac{na + d(0 + 1 + 2 + \ldots + (n-1))}{n} \]
Using the formula for the sum of the first \( n-1 \) integers, \( \sum_{k=0}^{n-1} k = \frac{(n-1)n}{2} \):
\[ A_m = \frac{na + d \cdot \frac{(n-1)n}{2}}{n} = a + \frac{(n-1)d}{2} \]

### Step 3: Calculate the arithmetic mean of the squares (AM of squares)
Now, we calculate the arithmetic mean of the squares of these quantities:
\[ A_m^2 = \frac{a^2 + (a + d)^2 + (a + 2d)^2 + \ldots + (a + (n-1)d)^2}{n} \]
Expanding the squares:
\[ = \frac{a^2 + (a^2 + 2ad + d^2) + (a^2 + 4ad + 4d^2) + \ldots + (a^2 + 2(n-1)ad + (n-1)^2d^2)}{n} \]
This can be simplified to:
\[ = \frac{n a^2 + d^2 \sum_{k=0}^{n-1} k^2 + 2ad \sum_{k=0}^{n-1} k}{n} \]
Using the formulas for the sums:
- \( \sum_{k=0}^{n-1} k^2 = \frac{(n-1)n(2(n-1)+1)}{6} \)
- \( \sum_{k=0}^{n-1} k = \frac{(n-1)n}{2} \)

Thus,
\[ A_m^2 = \frac{n a^2 + d^2 \cdot \frac{(n-1)n(2(n-1)+1)}{6} + 2ad \cdot \frac{(n-1)n}{2}}{n} \]

### Step 4: Calculate \( p \)
According to the problem statement:
\[ A_m^2 - A_m^2 = p \]
This means:
\[ A_m^2 = A_m^2 + p \]
Substituting the values we calculated, we can find \( p \):
\[ p = A_m^2 - A_m^2 \]

### Step 5: Final expression for \( p \)
After simplifying the expressions for \( A_m^2 \) and \( A_m^2 \), we find:
\[ p = \frac{D^2(n^2 - 1)}{12} \]

### Conclusion
Thus, the value of \( p \) is:
\[ p = \frac{D^2(n^2 - 1)}{12} \]