Asked by liza
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
Answers
Answered by
MathMate
For a geometric series,
the nth term is
t(n)=ar<sup>n</sup>
So
t(3)=ar³
t(6)=ar<sup>6</sup>
t(6)/t(3)
= ar<sup>6</sup> / ar³
= r<sup>6-3</sup>
= r³
Solve for r in
r³=9375/(-75)
r=(-125)<sup>1/3</sup>
=-5
A = t(3)/r³
= -75/(-5)³
= 3/5
and first term
= t(1)
= Ar
= 3/5(-5)
= -3
the nth term is
t(n)=ar<sup>n</sup>
So
t(3)=ar³
t(6)=ar<sup>6</sup>
t(6)/t(3)
= ar<sup>6</sup> / ar³
= r<sup>6-3</sup>
= r³
Solve for r in
r³=9375/(-75)
r=(-125)<sup>1/3</sup>
=-5
A = t(3)/r³
= -75/(-5)³
= 3/5
and first term
= t(1)
= Ar
= 3/5(-5)
= -3
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