There are 26 tuning forks arranged in the decreasing order of their frequencies. Each tuning fork gives 3 beats with the next. The first one is octave of the last. What is the frequency of 18th tuning fork ?

Options

A. 100
B. 99
C. 96
D. 103

Answer

To find the frequency of the 18th tuning fork, we can follow these steps:

### Step-by-Step Solution:

1. **Define the last tuning fork's frequency**:
Let the frequency of the last tuning fork (26th tuning fork) be \( N \).

2. **Determine the first tuning fork's frequency**:
Since the first tuning fork is an octave of the last, its frequency will be:
\[ f_1 = 2N \]

3. **Establish the relationship between the frequencies**:
The tuning forks are arranged in decreasing order, and each tuning fork gives 3 beats with the next. Therefore, the frequency of each subsequent tuning fork decreases by 3 Hz. The frequencies can be expressed as:
- \( f_1 = 2N \)
- \( f_2 = 2N - 3 \)
- \( f_3 = 2N - 6 \)
- ...
- \( f_k = 2N - 3(k-1) \)

4. **Find the frequency of the last tuning fork**:
For the 26th tuning fork, we have:
\[ f_{26} = 2N - 3(26-1) = 2N - 75 \]
Since \( f_{26} = N \), we can set up the equation:
\[ N = 2N - 75 \]

5. **Solve for \( N \)**:
Rearranging the equation gives:
\[ 75 = 2N - N \implies N = 75 \text{ Hz} \]

6. **Calculate the frequency of the 18th tuning fork**:
Now, we can find the frequency of the 18th tuning fork using the formula:
\[ f_{18} = 2N - 3(18-1) = 2N - 51 \]
Substituting \( N = 75 \):
\[ f_{18} = 2(75) - 51 = 150 - 51 = 99 \text{ Hz} \]

### Final Answer:
The frequency of the 18th tuning fork is **99 Hz**.

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