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A set of 20 tuning forks is arranged in a series of increasing frequencies. If each fork gives 4 beats with respect to the preceding fork and the frequency of the last fork is twice the frequency of the first, then the frequency of last fork is_____Hz.

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answer is 152.

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Detailed Solution

f₁ = f
f₂ = f + 4
f₃ = f + 2 × 4
f₄ = f + 3 × 4
f₂₀ = f + 19 × 4
f + (19 × 4) = 2 × f
f = 76 Hz.
Frequency of last tuning forks = 2f
= 152 Hz

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