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the third term of a geometric sequence is t3= -75 and the sixth term is t6= 9375. determine the first term and the common ratio

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For a geometric series,
the nth term is
t(n)=arn

So
t(3)=ar³
t(6)=ar6
t(6)/t(3)
= ar6 / ar³
= r6-3
= r³

Solve for r in
r³=9375/(-75)
r=(-125)1/3
=-5

A = t(3)/r³
= -75/(-5)³
= 3/5

and first term
= t(1)
= Ar
= 3/5(-5)
= -3

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