A string is fixed at both ends and transverse oscillations with amplitude  $a_0$ are excited.  Which of the following statements are correct?

If a string of length $l$  has cross-sectional area  $A,$ density of its material $\rho$  then its oscillation energy is given by
$E={\pi }^{2}A\rho a_0^{2}l{f}^2$
Where $f$  is frequency of transverse stationary wave formed in the string.
But  $f=\frac{v}{\lambda }=\frac{1}{\lambda }\sqrt{\frac{T}{m}}$
Where  $\lambda$ is wavelength, $T$  is tension in the string and  $m=A\rho .$
Since, string has a fixed length, therefore, wavelength of a tone excited in the string is constant.  Hence, energy  $E\propto T.$
Therefore, option (1) is correct.
If the frequency of fundamental tone is  $f_0,$ then frequency of  $n^{th}$ overtone will be equal to  $(n+1)f_0.$
Hence, oscillation energy of the string will be equal to :
$E_n={\pi }^{2}A\rho a_0^{2}l{f}_0^2{(n+1)}^2$
Since,   is not directly proportional to$n^2,$   therefore, option (2) is wrong.
Since, every particle of the performs SHM, therefore, r.m.s. speed of a particle $=1/\sqrt{2}\times$  its maximum speed
Hence, average KE is half of maximum KE.  But maximum KE is equal to oscillation energy of the string.  Therefore, option (3) is correct.