16 tuning forks are arranged in the order of | vedclass

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(A) Let the frequency of the first tuning fork be $n_1 = n$.Since there are $16$ tuning forks and each successive fork gives $8$ beats per second,the

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$120$

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(A) Let the frequency of the first tuning fork be $n_1 = n$.Since there are $16$ tuning forks and each successive fork gives $8$ beats per second,the frequencies form an arithmetic progression $(AP)$ with common difference $d = 8$ Hz.The number of terms is $N = 16$.The frequency of the last $(16^{th})$ tuning fork is given by $n_{16} = n_1 + (N - 1)d$.Substituting the values: $n_{16} = n + (16 - 1) 	imes 8 = n + 15 	imes 8 = n + 120$.According to the problem,the frequency of the last fork is twice the first,so $n_{16} = 2n$.Equating the two expressions: $2n = n + 120$.Solving for $n$: $n = 120$ Hz.

$16$ tuning forks are arranged in the order of increasing frequencies. Any two successive forks give $8$ beats per second when sounded together. If the frequency of the last fork is twice the first,then the frequency of the first fork is:

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