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