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Geometric progression

  • #1
just a check of my work please.

I have to write an expression for the nth term of this geometric sequence.

a1=100
a2=106
a3=112.36

I've worked out the ratio to be r=1.06

I am using the formula un=ar(n-1)

so the expression i have come up with is un=100(1.06)(n-1)

Is this correct? I have checked and it does work. I tried multiplying 100(1.06) on its own first to give me the formula
106(n-1)
but that doesnt work....hence me doubting myself.

Any comments much appreciated
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
hi ghostbuster25! :wink:
… so the expression i have come up with is un=100(1.06)(n-1)

Is this correct? I have checked and it does work. I tried multiplying 100(1.06) on its own first to give me the formula
106(n-1)
but that doesnt work....hence me doubting myself.
yup, un=100(1.06)(n-1) is fine :smile:

(i don't understand how you got 106(n-1) :confused:)
 
  • #3
Mentallic
Homework Helper
3,798
94
ghostbuster, if what you were trying to do is convert [tex]100(1.06)^{n-1}[/tex] into [tex]106^{n-2}[/tex], this is wrong because what you should be doing is taking out the factor of 1.06 to give [tex]100(1.06)(1.06)^{(n-1)-1}=106(1.06)^{n-2}[/tex].

106n grows much faster than 1.06n.
 

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