Find two fractions that are evenly spaced from 3/a and 7/2a.
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