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(A) Let the frequency of the first tuning fork be $n$.Since each fork produces $5 \, 	ext{beats/sec}$ with the next, the frequencies form an arithmet

Vedclass · Accepted Answer

$200, 400$

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(A) Let the frequency of the first tuning fork be $n$.Since each fork produces $5 \, 	ext{beats/sec}$ with the next, the frequencies form an arithmetic progression $(AP)$ with common difference $d = 5 \, 	ext{Hz}$.The number of tuning forks is $N = 41$.The frequency of the $N^{th}$ (last) tuning fork is given by the formula for the $N^{th}$ term of an $AP$: $l = a + (N - 1)d$.Given that the last frequency is double the first, $l = 2n$.Substituting the values: $2n = n + (41 - 1) 	imes 5$.$2n = n + 40 	imes 5$.$2n = n + 200$.$n = 200 \, 	ext{Hz}$.Therefore, the frequency of the first tuning fork is $200 \, 	ext{Hz}$ and the frequency of the last tuning fork is $2n = 400 \, 	ext{Hz}$.

$41$ tuning forks are arranged in increasing order of frequency. Each tuning fork produces $5 \, \text{beats/sec}$ with the next one. If the frequency of the last tuning fork is double that of the first, what are the frequencies of the first and last tuning forks?

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