For the string fixed on only one end, there is always an anti node at one end and a node at the other end. So, the length of the string gets divided into $1 / 4 t h$ of the wavelength $(\lambda) .$ The general formula for the length of the string is $L^{\prime}=n \lambda / 4$

The frequency $f$ becomes $V / \lambda, V$ is the velocity. In the first case, frequency $f=n V / 2 L$ $n=1,2,3, \dots$

In the second case, it is $n V / 4 L, \quad n=1,3,5,7 \ldots \ldots$ because of the length of the string will always have a half wave present. This makes $n$ an odd number.

For the first case$:$

$f_{1}=1 / 2 L(\sqrt{(T / \mu)})=50 H z=V / 2 \times L$

$f_{2}=2 \times f_{1}=100 H z=V / L$

$f 3=3 \times f_{1}=150 H z=3 V / 2 \times L$

Second case$:$

$n_{1}=V / 4 L$

$n_{2}=3 V / 4 L=\left(f-1+f_{2}\right) / 2=(100+50) / 2=75 H z$