Forty - one forks are so arranged that each products ` 5 beat//s` when sounded with its near fork . If the frequency of last fork is double the frequency of first and last fork , respectively are

Options

A. `200,400`
B. `205,410`
C. `195,390`
D. `100,200`

Explanation

To solve the problem, we need to analyze the arrangement of the forks and their frequencies. Let's break it down step by step.

### Step 1: Understand the arrangement of forks
We have 41 forks, and each fork produces 5 beats per second when sounded with its nearest fork. This means that the frequency of each fork differs from its neighboring fork by 5 Hz.

### Step 2: Define the frequencies
Let the frequency of the first fork be \( f_0 \). Then, the frequencies of the forks can be expressed as follows:
- Frequency of the 1st fork: \( f_0 \)
- Frequency of the 2nd fork: \( f_0 + 5 \)
- Frequency of the 3rd fork: \( f_0 + 10 \)
- ...
- Frequency of the 41st fork: \( f_0 + 40 \times 5 = f_0 + 200 \)

### Step 3: Establish the relationship between the first and last fork
According to the problem, the frequency of the last fork (41st fork) is double the frequency of the first fork. Therefore, we can set up the equation:
\[ f_0 + 200 = 2f_0 \]

### Step 4: Solve for \( f_0 \)
Rearranging the equation gives:
\[ 200 = 2f_0 - f_0 \]
\[ 200 = f_0 \]

### Step 5: Calculate the frequency of the last fork
Now that we have \( f_0 \), we can find the frequency of the last fork:
\[ \text{Frequency of the last fork} = f_0 + 200 = 200 + 200 = 400 \text{ Hz} \]

### Conclusion
The frequency of the first fork is \( 200 \) Hz and the frequency of the last fork is \( 400 \) Hz.

### Final Answer
The frequencies of the first and last forks are \( 200 \) Hz and \( 400 \) Hz, respectively.

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