Posted by Jon on Wednesday, December 12, 2007 at 11:15am.
1)Find the sixth term of the geometric sequence for which a1=5 and r=3
A)1215
B)3645
C)9375
D)23
I chose A
2)Write an equation for the nth term of the geometric sequence 12,4,4/3...
A)aN=12(1/3)n1
B)aN=12(1/3)n1
C)aN=12(1/3)n+1
D)aN=12(1/3)n1
I chose C
3)Find four geometric means between 5 and 1215.
A)+15,45,+135,405
B)15,45,135,405
C)247,489,731,973
D)+247,489,+731,973
I chose D
4)Find the sum of the geometric series 12864+32______to 8 terms
A)85
B)255
C)86
D)85/2
I chose A
5)Sigma again.
above the sigma sign there is a 6 and below is n=1.To the right is 5(4)n1. The question is find.
A)6825
B)4095
C)1023
D)5120
I chose B

math  Reiny, Wednesday, December 12, 2007 at 2:46pm
All are correct except #2 and #3
#2:
a+12, r = 1/3)
so it has to be D, I don't understand why you would choose C, the exponent does not follow the formula.
#3:
the sequence would be 5, 5r, 5r^2, 5r^3, 5r^4, 1215
so from the last 3 terms:
1215/5r^4 = 5r^4/5r^3
.
.
.
r^5=243
r=+3
so the terms in between would be 15, 45, 135, and 450 which is B

math  Jon, Wednesday, December 12, 2007 at 2:50pm
yeah for #2 i did pick D sorry for the typo
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