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We are given a frequency distribution table and asked to find the value of \( a \) given that the arithmetic mean is 5.

### Frequency Distribution Table:

| Variable (x) | 2   | 4   | 6   | 8     |
|--------------|-----|-----|-----|-------|
| Frequency (f)| \( a - 1 \) | \( a + 1 \) | \( a + 1 \) | \( 2a - 5 \) |

### Step-by-Step Solution:
The formula for the arithmetic mean of a frequency distribution is:

\[
\text{Arithmetic Mean} = \frac{\sum f_i x_i}{\sum f_i}
\]

Where:
- \( f_i \) are the frequencies,
- \( x_i \) are the variables.

1. **Calculate the sum of \( f_i x_i \) (weighted sum):**
   \[
   \sum f_i x_i = (a - 1) \times 2 + (a + 1) \times 4 + (a + 1) \times 6 + (2a - 5) \times 8
   \]

   Expanding:
   \[
   (a - 1) \times 2 = 2a - 2
   \]
   \[
   (a + 1) \times 4 = 4a + 4
   \]
   \[
   (a + 1) \times 6 = 6a + 6
   \]
   \[
   (2a - 5) \times 8 = 16a - 40
   \]

   Adding these up:
   \[
   \sum f_i x_i = (2a - 2) + (4a + 4) + (6a + 6) + (16a - 40)
   \]
   \[
   \sum f_i x_i = 2a + 4a + 6a + 16a - 2 + 4 + 6 - 40
   \]
   \[
   \sum f_i x_i = 28a - 32
   \]

2. **Calculate the sum of frequencies \( \sum f_i \):**
   \[
   \sum f_i = (a - 1) + (a + 1) + (a + 1) + (2a - 5)
   \]

   Simplifying:
   \[
   \sum f_i = a - 1 + a + 1 + a + 1 + 2a - 5
   \]
   \[
   \sum f_i = 5a - 4
   \]

3. **Substitute into the formula for the arithmetic mean:**
   We know the arithmetic mean is 5, so:
   \[
   5 = \frac{28a - 32}{5a - 4}
   \]

4. **Solve the equation:**
   Cross-multiply:
   \[
   5(5a - 4) = 28a - 32
   \]
   \[
   25a - 20 = 28a - 32
   \]

   Simplifying:
   \[
   -20 + 32 = 28a - 25a
   \]
   \[
   12 = 3a
   \]
   \[
   a = 4
   \]

Thus, the value of \( a \) is \( \boxed{4} \).

Would you like more details, or do you have any questions?

### Related Questions:
1. What is the arithmetic mean formula for grouped data?
2. How does changing \( a \) affect the frequency distribution?
3. What is the difference between mean and median in a frequency distribution?
4. How would the solution change if the arithmetic mean were a different value?
5. Can you explain the concept of weighted averages in other contexts?

### Tip:
Always ensure that you expand terms carefully when calculating weighted sums in frequency distributions to avoid errors in simplifying the final equation.

The arithmetic mean of the following frequency distribution is 5. Find the value of 'a'.
Variable: 2, 4, 6, 8
Frequency: a-1, a+1, a+1, 2a-5

Learn how to calculate the value of 'a' in a frequency distribution table where the arithmetic mean is given as 5. This problem covers algebraic manipulation and the formula for the arithmetic mean in frequency distributions, suitable for students in Grades 10-12.

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