We are looking for two numbers, call them x and y such that:

(3/a) + d = x

x + d = y

y + d = 7/(2a)

where d is some constant.

Putting all these together we get that

3/a + 3d = 7/(2a)

1/a + d = 7/(6a)

d = 7/(6a) - 1/a = 7/(6a) - 6/(6a) = 1/(6a)

So

x = 3/a + d = 3/a + 1/(6a) = 18/(6a) + 1/(6a) = 19/(6a)

y = x + d = 19/(6a) + 1/(6a) = 20/(6a) = 10/(3a)

To check:

7/(2a) =? y + d = 20/(6a) + 1/(6a) = 21/(6a) = 7/(2a) Check!

So the two intermediate numbers are 19/(6a) and 10/(3a).

-Dan