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In a geometric series, the first term is $12$, the third term is $92$, and the sum of all of the terms of the series is $62 813$. How many terms are in the series?

According to the answer sheet of the pre-calculus 11 book, the number terms in the series is 13. Can anyone explain how does that happen?

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What have you tried? –  yohBS Feb 20 '13 at 22:49
    
By $\,a^1\,,\,a^3\,$ do you mean you mean indexes or powers of a constant number $\,a\,$? –  DonAntonio Feb 20 '13 at 23:00
    
No, this is wrong. It means it is the first term. I think I will change the question to make it more specific and correct. –  clixbrigidxterx Feb 20 '13 at 23:04
    
I see: someone incorrectly edited your question. This is one reason more why it is better to write from the beginning with LaTeX, as shown in the FAQ section... –  DonAntonio Feb 20 '13 at 23:08

2 Answers 2

use the formula for the sum of a geometric sequence that is if the sequence is $a^0+a^1+a^2+...a^n$ then the sum is $\frac{a^{n+1}-1}{a-1}.$ Therefore $$12\frac{\sqrt\frac{92}{12}^{n+1}-1}{\sqrt\frac{92}{12}-1}=62813$$ so $$\frac{\sqrt\frac{92}{12}^{n+1}-1}{\sqrt\frac{92}{12}-1}= \frac{62813}{12}$$then $$\sqrt{\frac{92}{12}}^{n+1}-1=62813\frac{\sqrt\frac{92}{12}-1}{12}$$ now pass the -1 to the other side and take the log to get. $$(n+1) \log \sqrt{\frac{92}{12}} =\log (62813\frac{\sqrt\frac{92}{12}-1}{12}+1)$$

and finally $$n=\frac{\log (62813\frac{\sqrt\frac{92}{12}-1}{12}+1)}{\log \sqrt{\frac{92}{12}}}-1$$

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So, what is the answer? –  clixbrigidxterx Feb 20 '13 at 23:22
    
That is the answer, now you have to check what integer it is equal to ( if it is an integer) –  Jorge Fernández Feb 20 '13 at 23:48
    
Is there other way without using logarithms? Our teacher taught us answering the geometric series problems without the use of logarithms. –  clixbrigidxterx Feb 20 '13 at 23:57
    
I think what the OP means is that $\sum^3 a^k =92$ and $\sum^1 a^k=12$; or am I reading "series" and he meant "progression" or "sequence"? –  Pedro Tamaroff Feb 20 '13 at 23:59
    
We are only in Geometric Sequence and Series. We are not yet in progression or infinite something. :/ –  clixbrigidxterx Feb 21 '13 at 0:03

Let the series be $a+ar+ar^2+\cdots +ar^{n-1}$. This has $n$ terms. Clearly $a=12$. We are told that $ar^2=92$, giving $r^2=\frac{92}{12}$, so $r=\sqrt{\frac{92}{12}}$. Unusually unattractive number for this level.

The sum $S$ of the series is $62813$. Not too pretty either! It will make things clearer, and typing easier, if we just continue with letters. The sum $S$ of the series is given by $$S=a\frac{r^n-1}{r-1}.$$ Solving for $r^n$, we obtain $$r^n=1+\frac{(r-1)S}{a}.$$ Let the right-hand side be $W$. Take your favourite log of both sides. We get $n=\dfrac{\log(W)}{\log(r)}$. I do not get $n=13$.

Possibilities: (i) I have made an error or (ii) someone else has made an error or (iii) there is a typo in the numbers of the question. The ugliness of the numbers makes it plausible that other numbers were intended. If that is the case, the formula above is ready for them.

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I just wish that the book is incorrect. :) About the common ratio, I got a 2.77 which is the closest when it is rounded off. It's just I do not know how to solve the final step when I am solving for the n. –  clixbrigidxterx Feb 21 '13 at 0:29
    
Yes, the common ratio is about $2.769$. Rounding when we are dealing with exponential growth can make a big difference. For the final step, once we know $r^n$, there are two ways. Logarithms, or experimentation. If you are solving $r^n=B$, where you know $r$ and $B$, you can guess at an $n$, check whether it works, adjust. The process can be pretty fast. –  André Nicolas Feb 21 '13 at 0:39
    
I really just don't know how they get 13 as number of terms. So, can you show me how can I solve this without using logarithms with a different answer? –  clixbrigidxterx Feb 21 '13 at 1:21
    
Thought I explained. From the formula I gave in the post, we have $2.769^{n}\approx 9260.68$. Guess $n=8$. Calculate $2.769^8$. About 3456, too small. Continue. –  André Nicolas Feb 21 '13 at 1:37

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