41 tuning forks are arranged in such a way th | vedclass

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(A) Let the frequency of the first tuning fork be $n_1 = n$.Since each fork produces $5 	ext{ beats per second}$ with the next,the frequencies form a

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$200 	ext{ Hz}, 400 	ext{ Hz}$

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(A) Let the frequency of the first tuning fork be $n_1 = n$.Since each fork produces $5 	ext{ beats per second}$ with the next,the frequencies form an arithmetic progression with common difference $d = 5 	ext{ Hz}$.The number of tuning forks is $N = 41$.The frequency of the $N^{th}$ (last) fork is given by $n_N = n_1 + (N - 1)d$.Substituting the values: $n_{41} = n + (41 - 1) 	imes 5 = n + 40 	imes 5 = n + 200$.Given that the frequency of the last fork is double the frequency of the first fork,we have $n_{41} = 2n$.Equating the two expressions: $2n = n + 200$.Solving for $n$: $n = 200 	ext{ Hz}$.Therefore,the frequency of the first fork is $n_1 = 200 	ext{ Hz}$ and the frequency of the last fork is $n_{41} = 2 	imes 200 = 400 	ext{ Hz}$.

$41$ tuning forks are arranged in such a way that each produces $5 \text{ beats per second}$ when sounded with its nearest fork. If the frequency of the last fork is double the frequency of the first fork,then the frequencies of the first and last fork are respectively:

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