In case of stationary wave pattern, which of the following statement is CORRECT ?

The distance between consecutive antinodes in a stationary wave is $\dfrac{\lambda}{2}$, not $\lambda$.

In a pipe open at both ends, all harmonics (both even and odd) are present, with frequencies $f_n = n \dfrac{v}{2L}$ where $n = 1, 2, 3, \dots$

In a pipe closed at one end, only odd harmonics are present, with frequencies $f_n = (2n-1) \dfrac{v}{4L}$ where $n = 1, 2, 3, \dots$

In a stretched string fixed at both ends, the frequencies of vibration are $f_n = n \dfrac{v}{2L}$. The fundamental frequency is $f_1$ (first harmonic). The first overtone corresponds to the next higher frequency, which is $f_2 = 2f_1$. This is exactly the second harmonic. Thus, the frequency of the first overtone is the same as the second harmonic.