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Given are the following geometric sequences: 13, 23.4, ...

The common ratio is 1.8, so far so good.

But how can I calculate the number of terms which are smaller then 9.6E13?

The solution says 51. I have no clue.

I'm looking for a hint to solve this. Thanks in advance.

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Can you figure out how big the $n$th term is? –  Max Morin Oct 21 '12 at 17:03
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2 Answers 2

up vote 1 down vote accepted

Hint:

$$13\cdot (1.8)^{n-1}\geq 9.6\times 10^{13}\Longrightarrow 1.8^{n-1}\geq 0.738461\times 10^{13}=:\alpha\Longrightarrow n-1\geq\frac{\log\alpha}{\log 1.8}=....$$

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Thanks...I got the solutions steps. Actually, it's other way with the sign. It should be smaller then 9.6E13. But thanks anyway. –  suhshio Oct 21 '12 at 19:37
    
I know you want to know the number of elements smaller than the given number, but knowing the first element bigger than it, which is the 52th, you automatically know that there are 51 elements less than that...so the sign is correct if one knows how to interpret it. –  DonAntonio Oct 22 '12 at 2:50
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Write it as $a_{n}=13\cdot 1.8^{n}$, $n\ge 0$. Then solve the inequality $a_{n}\ge 9.6E13$. That will tell you the first $n$ for which the terms arent smaller than 9.6E13, and it should be simple to find the answer from there

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I think you meant $\,a_n=13\cdot (1.8)^{n-1}\,$ ... –  DonAntonio Oct 21 '12 at 17:25
    
@DonAntonio I think you're right about that. –  suhshio Oct 21 '12 at 17:28
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@Peter In my opinion it leads to an logarithmic calculation then, right? –  suhshio Oct 21 '12 at 17:29
    
Indeed it does, @swisshenry –  DonAntonio Oct 21 '12 at 17:30
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