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)n-1

B)aN=12(-1/3)n-1

C)aN=-12(-1/3)-n+1

D)aN=-12(-1/3)n-1

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 128-64+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)n-1. 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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