A set of 28 tuning forks is arranged in an in | vedclass

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(B) Let the frequencies of the $28$ tuning forks be $n_1, n_2, \dots, n_{28}$ in increasing order.Since each fork produces '$x$' beats per second with

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$4$

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(B) Let the frequencies of the $28$ tuning forks be $n_1, n_2, \dots, n_{28}$ in increasing order.Since each fork produces '$x$' beats per second with the preceding one,the frequencies form an Arithmetic Progression $(AP)$ with common difference $d = x$.Thus,the frequency of the $k^{	ext{th}}$ fork is $n_k = n_1 + (k-1)x$.For the $12^{	ext{th}}$ fork: $n_{12} = n_1 + 11x = 152 	ext{ Hz} \quad \dots(1)$.For the $28^{	ext{th}}$ fork: $n_{28} = n_1 + 27x$.Given that the last fork is an octave of the first,$n_{28} = 2n_1$.Substituting the expression for $n_{28}$: $2n_1 = n_1 + 27x \Rightarrow n_1 = 27x$.Substitute $n_1 = 27x$ into equation $(1)$:$27x + 11x = 152$$38x = 152$$x = \frac{152}{38} = 4 	ext{ Hz}$.

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