First, an example:
Suppose the frames are 1500 bytes long, it is the same as 12000 bits.
The transmission of the frame to A at 10 Mb/s takes 1.2 milliseconds.
The transmission of the same size frame to B takes 6 milliseconds.
In total, to transmit 24000 bits took 7.2 milliseconds. It gives an speed of 3.33 Mb/s.
The result isn't the arithmetic mean, is the Harmonic Mean.
The harmonic mean can be expressed as the reciprocal of the arithmetic mean of the reciprocals.
For our case:
=1/(((1/10)+(1/2))/2) =1/((6/10)/2) = (2/1)/(6/10) = 20/6 = 3.33
In general, for two speeds s1 and s2 will be:
=(2*s1*s2)/(s1+s2)
In this case =(2*10*2)/(10+2) =(40/12) =3,33