If the observation 1, 2, 3, ……….., n occur with frequency, `n,(n-1), (n-2),…..,1` respectively such that the mean of observations is `(13)/(3)`, then n is equal to

Options

A. 10
B. 11
C. 12
D. 13

Answer

To solve the problem, we need to find the value of \( n \) given that the observations \( 1, 2, 3, \ldots, n \) occur with frequencies \( n, n-1, n-2, \ldots, 1 \) respectively, and the mean of the observations is \( \frac{13}{3} \).

### Step-by-Step Solution:

1. **Define the Observations and Frequencies**:
- Observations: \( x_i = 1, 2, 3, \ldots, n \)
- Frequencies: \( f_i = n, n-1, n-2, \ldots, 1 \)

2. **Calculate the Mean**:
The mean \( \bar{x} \) is given by the formula:
\[ \bar{x} = \frac{\sum (x_i \cdot f_i)}{\sum f_i} \]

3. **Calculate \( \sum f_i \)**:
The sum of frequencies is:
\[ \sum f_i = n + (n-1) + (n-2) + \ldots + 1 = \frac{n(n+1)}{2} \]

4. **Calculate \( \sum (x_i \cdot f_i) \)**:
We need to compute:
\[ \sum (x_i \cdot f_i) = 1 \cdot n + 2 \cdot (n-1) + 3 \cdot (n-2) + \ldots + n \cdot 1 \]
This can be simplified as:
\[ = n + 2(n-1) + 3(n-2) + \ldots + n \cdot 1 \]
Rearranging gives:
\[ = n + 2n - 2 + 3n - 6 + \ldots + n^2 \]
This can be expressed as:
\[ = n \cdot \sum_{i=1}^{n} i - \sum_{i=1}^{n} i^2 \]
Using the formulas:
\[ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \quad \text{and} \quad \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \]
We get:
\[ \sum (x_i \cdot f_i) = n \cdot \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6} \]

5. **Combine the Results**:
Substitute back into the mean formula:
\[ \frac{n \cdot \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}} = \frac{13}{3} \]
Simplifying gives:
\[ \frac{3n^2 + 3n - n(2n+1)}{3n} = \frac{13}{3} \]

6. **Clear the Denominator**:
Multiply both sides by \( 3n \):
\[ 3n^2 + 3n - 2n^2 - n = 13n \]
Rearranging leads to:
\[ n^2 - 11n = 0 \]

7. **Factor the Equation**:
\[ n(n - 11) = 0 \]
Thus, \( n = 0 \) or \( n = 11 \). Since \( n \) must be a positive integer, we have:
\[ n = 11 \]

### Final Answer:
Thus, the value of \( n \) is \( 11 \).