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A geometric sequence has 400 terms.  The first term is 1000 and the common ratio is -\frac{9}{10}.  How many terms of this sequence are greater than 1?

 Feb 7, 2024
 #1
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We need to solve this

 

1000*(9/10)^(n-1) =  1

 

(.9)^(n -1)  =   1/1000       take the log of  both sides

 

log (.9)^(n-1) = log (1/1000)

 

And we can write

 

(n-1) log (.9)  = log (1/1000)

 

n -1  =  log (1/1000) / log (.9)

 

n -1   = 65.56

 

n = 66.56

 

 

The 66th term is  1000(.9)(66-1)  = 1.061

The 67th term is  1000(.9)(67-1) = .955

 

So  66 terms are > 1

 

cool cool cool

 Feb 7, 2024

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