Hi,
Since the first term of a GP is more than that of G.P.'s second or third term, the common ratio must be negative. (It is given that a is median,which implies a is definitely more than either b or c since a, b, and c
are distinct).
It is given that the sum of these three terms is 42 or -42, and the r is a negative integer.
$$a\left(1+r+r^2\right)=\pm\ 42$$. Now 42 can be factorized as (1*42), (2*21), (3*14), (6*7).
Now the value of $$1+r+r^2\ $$ can be 1/2/3/6/7/14/21/42.
When $$1+r+r^2\ $$ is 1, then the value of a can be +42 or -42.

$$1+r+r^2\ =1\ =>\ r\left(r+1\right)=0\ =>\ r=0\ or\ -1$$(r can't be zero)
as r must be a negative integer.
If r=-1, then the terms will be (+42, -42, 42 ) or (-42, 42, -42), which is not possible since a, b, and c must be distinct.
Similarly, if $$1+r+r^2\ $$ =2, then r is not an integer. If $$1+r+r^2\ $$=3, then a value can be +14 or -14.
$$1+r+r^2\ =\ 3=>\ r^2+r-2=0\ =>\ \left(r+2\right)\left(r-1\right)=0\ =>\ r\ =\ -2\ or\ 1$$ ( r can't be positive).
If r= -2, then the terms will be (+14, -28, 56) or (-14, 28, -56).
Similarly, if $$1+r+r^2\ $$=6, then r is not an integer. If $$1+r+r^2\ $$ = 7, then the value of a is either 6 or -6.
$$1+r+r^2\ =\ 7=>\ r^2+r-6=0\ =>\ \left(r+3\right)\left(r-2\right)=0\ =>\ r\ =\ -3\ or\ 2$$ ( r can't be positive).
When r = -3, the terms will be (6, -18, 54) or (-6, 18, -54).
Similarly, if $$1+r+r^2\ $$=14, then r is not an integer. If $$1+r+r^2\ $$=21, then the value of a can be +2 or -2.
$$1+r+r^2\ =\ 21=>\ r^2+r-20=0\ =>\ \left(r+5\right)\left(r-4\right)=0\ =>\ r\ =\ -5\ or\ 4$$ ( r can't be positive)
When r = -5, the terms will be (2, -10, 50) or (-2, 10, -50). Similarly, if $$1+r+r^2\ $$=42, then r is not an integer.
Therefore, the value b can take is -10/10/-18/18/-28/28.
I hope this helps!