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

Answer 1

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

So
t(3)=ar³
t(6)=ar⁶
t(6)/t(3)
= ar⁶ / ar³
= r^6-3
= 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