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.

We need to find the frequency of the last tuning fork in a series of 20 forks.

There are 20 tuning forks and each fork produces 4 beats per second with the preceding fork because the frequencies increase by a constant amount. The frequency of the last fork is given to be twice the frequency of the first fork.

Since the forks form an arithmetic progression with common difference $$d = 4$$ Hz, let the frequency of the first fork be $$f$$. Then the frequency of the $$n$$th fork is $$f + (n-1)\times4$$.

Now for the 20th fork, its frequency is $$f + 19 \times 4 = f + 76$$. Given that this equals $$2f$$, substituting yields $$f + 76 = 2f$$, which gives $$f = 76$$ Hz.

From this, the frequency of the last fork is $$f_{20} = 2f = 2 \times 76 = 152$$ Hz.

The frequency of the last fork is 152 Hz.