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Posted by 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 - , 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 - , Wednesday, December 12, 2007 at 2:50pm

    yeah for #2 i did pick D sorry for the typo

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