A set of $14$ tuning forks is arranged in a series of increasing frequencies. Each fork produces '$x$' beats per second with the preceding fork and the last fork is an octave of the first fork. If the seventh fork has the frequency of $114$ Hz, the value of $x$ is

Let the frequency of the first tuning fork be $f_1$.

Since each fork produces $x$ beats per second with the preceding fork and the frequencies are in increasing order, the frequencies form an arithmetic progression with a common difference of $x$.

The frequency of the $n$-th tuning fork is given by $f_n = f_1 + (n - 1)x$.

The frequency of the 14th tuning fork is $f_{14} = f_1 + 13x$.

It is given that the last fork is an octave of the first fork, which means $f_{14} = 2f_1$.

Substituting this into the equation, we get $2f_1 = f_1 + 13x \Rightarrow f_1 = 13x$.

The frequency of the 7th tuning fork is given as $114$ Hz.

Using the formula for the 7th fork, $f_7 = f_1 + 6x$.

Substituting $f_1 = 13x$, we get $f_7 = 13x + 6x = 19x$.

Equating this to the given frequency, $19x = 114$.

Solving for $x$, we get $x = \dfrac{114}{19} = 6$.

Answer: $6$