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