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Let the frequencies of the tuning forks are in AP with common difference -x (since they're arranged in decreasing order of frequency).
Let the frequency of first tuning fork be,
a(1) = n
So the frequency of p'th tuning fork is,
a(p) = n - (p - 1)x
Thus the frequency of the last (33rd) tuning fork is,
a(33) = n - 32x → (1)
We have the frequency of the 10th tuning fork,
a(10) = 440
n - 9x = 440
n = 9x + 440 → (2)
Given that first fork is octave of the last. It means the frequency of the first fork is double that of the last fork. Hence,
a(1) = 2 · a(33)
n = 2(n - 32x)
n = 2n - 64x
n = 64x
9x + 440 = 64x [From (2)]
x = 8 Hz
Thus the frequency of the first fork is, from (2),
a(1) = 9 × 8 + 440
a(1) = 512 Hz
And that of the last fork is, from (1),
a(33) = 512 - 32 × 8
a(33) = 256 Hz
[tex] \mathfrak{ \huge{ \red{ \underline{ \underline{Answer}}}}}[/tex]
=> A set of 33 tuning fork is arranged in series with decreasing order of frequency.
=> Each fork produces x beats per second with previous one.
=> Freq. of first fork is double than freq. of last fork.
=> Freq. of 10th fork is 440 Hz.
=> Freq. of first and last fork
=> let total number of forks = n
=> freq. of first fork = f1
=> freq. of 10th fork f10 = 440 Hz = f1 - 9x
=> freq. of 33th fork f33 = f1 - 32x
=> f33 = f1/2 therefore f1 = 64x
=> 440 = 64x - 9x = 55x
=> x = no. of beats = 440/55 = 8
=> Freq. of first fork = 64x = 512Hz
=> Freq. of last fork = 512 - 32(8) = 256Hz
[tex] \boxed{ \boxed{ \blue{ \bold{ \sf{ \huge{f(1) = 512 \: Hz \: and \: f(33) = 256 \: Hz}}}}}}[/tex]