1. ## Limiting Sums

Hey could someone help me with this question?
The third term of a geometric series is 54 and the sixth therm is 11 and 83/125. Evaluate a and r and its limiting sum.

Thanks

2. ## Re: Limiting Sums

Originally Posted by civilKRobbo
Hey could someone help me with this question?
The third term of a geometric series is 54 and the sixth therm is 11 and 83/125. Evaluate a and r and its limiting sum.

Thanks
What is the 83/125?

3. ## Re: Limiting Sums

Originally Posted by civilKRobbo
Hey could someone help me with this question?
The third term of a geometric series is 54 and the sixth therm is 11 and 83/125. Evaluate a and r and its limiting sum.

Thanks
When someone poses an incompletely-specified question it shows he/she does not fully understand the question. Often a key part of the question is omitted, he/she being unaware it is an important part of the whole question.

4. ## Re: Limiting Sums

Originally Posted by 1729
What is the 83/125?
Originally Posted by Drongoski
When someone poses an incompletely-specified question it shows he/she does not fully understand the question. Often a key part of the question is omitted, he/she being unaware it is an important part of the whole question.
$\noindent Possibly, the original poster meant 11\tfrac{83}{125}.$

5. ## Re: Limiting Sums

Originally Posted by InteGrand
$\noindent Possibly, the original poster meant 11\tfrac{83}{125}.$
That makes sense. You are a great decipherer.
In that case:

$ar^2 = 54 and ar^5 = 11\frac {83}{125} = \frac {1458}{125}\\ \\ \therefore \frac {ar^5}{ar^2} = r^3 = 0.126 \implies r = 0.6 and a = 150 \implies S_{\infty} = \frac {a}{1-r} = \frac {150}{1-0.6} = 375$

6. ## Re: Limiting Sums

Originally Posted by civilKRobbo
Hey could someone help me with this question?
The third term of a geometric series is 54 and the sixth therm is 11 and 83/125. Evaluate a and r and its limiting sum.

Thanks
Originally Posted by Drongoski
That makes sense. You are a great decipher.
$\noindent Assuming this was the case, the third term is ar^{2} = 54 and the sixth term is ar^{5} = 11\tfrac{83}{125}. This implies that r^{3} = \frac{ar^{5}}{ar^{3}} = \frac{11\tfrac{83}{125}}{54} = \frac{27}{125}. Taking cube roots, we have \boxed{r = \frac{3}{5}}. Now$

\begin{align*}ar^{2} &= 54 \\ \Rightarrow a &= \frac{54}{r^{2}}\\ \Rightarrow \boxed{a = 150}. \quad \Big{(}\text{as } r= \frac{3}{5}\Big{)}\end{align*}

$\noindent The limiting sum is thus \frac{a}{1 - r} = \frac{150}{1 - \frac{3}{5}} = 375.$

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