Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 2 down vote favorite

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.

share|improve this question
Can you figure out how big the $n$th term is? – Max Morin Oct 21 '12 at 17:03

2 Answers

up vote 1 down vote accepted

Hint:

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

share|improve this answer
Thanks...I got the solutions steps. Actually, it's other way with the sign. It should be smaller then 9.6E13. But thanks anyway. – devhenry 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
up vote 2 down vote

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

share|improve this answer
1  
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. – devhenry Oct 21 '12 at 17:28
1  
@Peter In my opinion it leads to an logarithmic calculation then, right? – devhenry Oct 21 '12 at 17:29
Indeed it does, @swisshenry – DonAntonio Oct 21 '12 at 17:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.