1. Let the frequencies of the 20 tuning forks be in increasing order:

$f_1, f_2, f_3, \dots, f_{20}$

Since each fork gives 4 beats with respect to the preceding fork, the difference between successive frequencies is:

$f_{n+1} - f_n = 4\,\text{Hz}$

So the frequencies are in an arithmetic progression (A.P.) with common difference:

$d = 4$

2. Therefore,

$f_{20} = f_1 + 19\times 4 = f_1 + 76$

3. It is given that the frequency of the last fork is twice the frequency of the first:

$f_{20} = 2f_1$

Substitute from above:

$f_1 + 76 = 2f_1$

$f_1 = 76\,\text{Hz}$

4. Hence,

$f_{20} = 2f_1 = 2\times 76 = 152\,\text{Hz}$

5. Final answer:

$\boxed{152\,\text{Hz}}$

The derived answer matches the stored correct answer.