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Geometric progression

  • Thread starter nae99
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  • #1
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Homework Statement



3,6,12........1536
determine the number of terms in the progression

Homework Equations





The Attempt at a Solution


a=3 r=2

n= ar^n-1

1536= (3) (2)^n-1
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
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Okay, so solve it! First, divide both sides by 3. If that sequence is actually geometric, you should be able to identify 1536/3 as a power of 2. I suggest you just calculate powers of 2: 1, 2, 4, 8, 16, ... until you get to that number.
 
  • #3
129
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Okay, so solve it! First, divide both sides by 3. If that sequence is actually geometric, you should be able to identify 1536/3 as a power of 2. I suggest you just calculate powers of 2: 1, 2, 4, 8, 16, ... until you get to that number.
1536= (3) (2)^n-1

1536/3 = 6^n-1[itex]/[/itex]3

512 = 2^n-1

512 = 2^10-1

512 = 2^9

n = 10
 
  • #4
33,641
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1536= (3) (2)^n-1

1536/3 = 6^n-1[itex]/[/itex]3

512 = 2^n-1

512 = 2^10-1

512 = 2^9

n = 10
There are errors in your work. Also, you need parentheses around your exponent expressions.
1536= (3) (2)^(n-1)

1536/3 = 6^n-1[itex]/[/itex]3
The above is incorrect. 3*2^(n - 1) [itex]\neq[/itex] 6^(n - 1)
512 = 2^n-1
The above is also incorrect. [6^(n - 1)]/3 [itex]\neq[/itex] 2^(n - 1)
512 = 2^10-1

512 = 2^9

n = 10
 
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