5 members of a team are weighed consecutively and their average weight calculated after each member is weighed. If the average weight increases by one kg each time, how much heavier is the last player than the first one ?

Options

A. 4 kg
B. 20 kg
C. 8 kg
D. 5 kg

Answer

To solve the problem, we need to analyze the information given about the weights of the team members and how the average weight changes as each member is weighed.

### Step-by-Step Solution:

1. **Understanding the Average Weight**:
- Let the weight of the first member be \( w_1 \).
- The average weight after weighing the first member is \( w_1 \).

2. **Calculating the Average After Each Member**:
- When the second member is weighed, let their weight be \( w_2 \).
- The average weight after weighing the second member is:
\[ \text{Average after 2 members} = \frac{w_1 + w_2}{2} \]
- According to the problem, this average increases by 1 kg compared to the average after the first member:
\[ \frac{w_1 + w_2}{2} = w_1 + 1 \]
- Rearranging gives:
\[ w_1 + w_2 = 2w_1 + 2 \implies w_2 = w_1 + 2 \]

3. **Continuing the Process**:
- Let the weight of the third member be \( w_3 \).
- The average after weighing the third member is:
\[ \frac{w_1 + w_2 + w_3}{3} \]
- This average should be 1 kg more than the average after weighing the second member:
\[ \frac{w_1 + w_2 + w_3}{3} = w_1 + 2 + 1 = w_1 + 3 \]
- Substituting \( w_2 = w_1 + 2 \):
\[ \frac{w_1 + (w_1 + 2) + w_3}{3} = w_1 + 3 \]
- Simplifying gives:
\[ \frac{2w_1 + 2 + w_3}{3} = w_1 + 3 \]
- Multiplying through by 3:
\[ 2w_1 + 2 + w_3 = 3w_1 + 9 \implies w_3 = w_1 + 7 \]

4. **Repeating for the Fourth and Fifth Members**:
- Let \( w_4 \) be the weight of the fourth member.
- The average after the fourth member is:
\[ \frac{w_1 + w_2 + w_3 + w_4}{4} = w_1 + 4 \]
- Substituting the known weights:
\[ \frac{w_1 + (w_1 + 2) + (w_1 + 7) + w_4}{4} = w_1 + 4 \]
- This simplifies to:
\[ \frac{3w_1 + 9 + w_4}{4} = w_1 + 4 \]
- Multiplying through by 4:
\[ 3w_1 + 9 + w_4 = 4w_1 + 16 \implies w_4 = w_1 + 7 \]

5. **Finding the Weight of the Fifth Member**:
- Let \( w_5 \) be the weight of the fifth member.
- The average after the fifth member is:
\[ \frac{w_1 + w_2 + w_3 + w_4 + w_5}{5} = w_1 + 5 \]
- Substituting the known weights:
\[ \frac{w_1 + (w_1 + 2) + (w_1 + 7) + (w_1 + 7) + w_5}{5} = w_1 + 5 \]
- This simplifies to:
\[ \frac{4w_1 + 16 + w_5}{5} = w_1 + 5 \]
- Multiplying through by 5:
\[ 4w_1 + 16 + w_5 = 5w_1 + 25 \implies w_5 = w_1 + 9 \]

6. **Calculating the Difference**:
- Now we have:
- \( w_1 \) (weight of the first member)
- \( w_5 = w_1 + 9 \) (weight of the last member)
- The difference in weight between the last and the first member is:
\[ w_5 - w_1 = (w_1 + 9) - w_1 = 9 \text{ kg} \]

### Final Answer:
The last player is **9 kg heavier** than the first player.