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 _____ $\mathrm{Hz}$.

Each fork produces $4$ beats per second with the previous means each fork has frequency $4 \mathrm{Hz}$ more than the previous.

Using relation, $f_{\mathrm{last}}=f_{\mathrm{first}}+\left(N-1\right)x$, here, $N$ is the number of tuning fork in series and $x$ is beat frequency between two successive forks.

$2f_0=f_0+ \left(20-1\right)4$

$$\begin{gathered} \Rightarrow 2f_0=f_0+76 \\ \Rightarrow f_0=76 \mathrm{Hz} \end{gathered}$$

Frequency of last tuning fork is $152 \mathrm{Hz}$.