Auniform narrow tube 1.80 m long is open at both ends. it resonates at two successive harmonics of frequencies 280 hz and 300 hz. (a) what is the fundamental frequency(hz)? (b) what is the speed of sound in the gas in the tube(m/s)?

 Harmonics are integer multiples of the fundamental frequency. 

I found the greatest common factor of 280 Hz and 420 Hz, and that was 140 Hz. Therefore, I knew that this was the fundamental frequency. (280 Hz and 420 Hz are the second and third harmonics, respectively) 

The formula for the fundamental frequency of an open pipe is 

f= (v/2L) 
where v is the velocity of the gas inside the tube and L is the length of the tube. 
I assume that I'm ignoring end corrections in this calculation. 

f= 140 Hz 
v= unknown 
L= 1.85 meters 

140 = (v)/ (2*1.85) 
v= (140)(2)(1.85)= 518 m/s

Ithink the answer is a