If the third and ninth temr of a geometric series with a positive common ratio are -3 and -192 repectively, determine the value of

the first term a
Let a_1 be denoted by 'a' (first terms) then,
a_3=r^2 a
And,
a_9=r^8 a
Thus,
r^8 a=-192
r^2 a=-3
Divide two equations,
r^6=64=2^6
Thus,
r=2

The third term is,
-3=2^2 a
Thus,
a=-3/4

You used the ratio of t_n/t_m = r^n-m, where n>m. In this case n=9 and m=3.
It looks like you're getting a good handle on geometric sequences. Well done!