I'm currently doing a Pre-Calculus refresher and am stuck on a question for an assignment. Maybe I am overthinking it, but I've been working on it for an hour, and rather than getting closer to an answer, I seem to be getting more confused.
Question is:
The sum of the infinite geometric series $t + t^2 + t^3 + t^4 + ... = 4t$, $t$ != $0$. Determine the value of $t$.
What I have worked out so far:
I know that I need to re-work the sum of infinite geometric series formula, which is $S\infty = \frac{a}{(1 - r)}$ in which $r < 1$
- I know that $S\infty = 4t$
- I know that $a = t$
- I need to determine $r$ (See below, works out to $r = 0.75$)
I have tried to work out the value of $r$ by re-working the sum of infinite geometric series formula as follows:
- $4t = \frac{t}{(1-r)}$
- $(1-r) * 4t = t$
- $(1-r) = \frac{t}{4t}$
- $(1-r) = \frac{1}{4}$
- $1 = \frac{1}{4} + r$
- $1 - \frac{1}{4} = r$
- $0.75 = r$
From here, I can plug in the value of $r$ into the sum of infinite geometric series formula:$$4t = \frac{t}{(1 - 0.75)}$$
This is the part where I become confused, because it seems that I need the value of $a$ to be able to solve for $t$. However, I do not have a value for $a$ as my first value in the given series is the unknown value $t$. How can I solve this equation with the information I have been given? Am I missing something?
Thank you in advance for any help given! I hope I have worded my question in a way that can be understood. :)
EDIT: My solution after everyone's help:
- $4t = \frac{t}{(1-t)}$
- $(1-t) * 4t = t$
- $4t - 4t^2 = t$
- $3t - 4t^2 = 0$
- $t(3 - 4t) = 0$
- divide each side by t, to get $3 - 4t = 0$
- $3 = 4t$
- $\frac{3}{4} = t$
Thanks so much to everyone who answered!
$\endgroup$ 32 Answers
$\begingroup$First term $a=t$ and common ratio $q=t$
$$S_{\infty} = \frac {a}{1-q}=\frac {t}{1-t}=4t,-1\leq q\leq 1$$$$4t=3,t=\frac {3}{4}$$
$\endgroup$ 2 $\begingroup$Since t isn't 0, divide by t. We get$1+t+t^2+t^3+t^4...=4$Simplify to get
- $t=3-(t^2+t^3+t^4...)$
We know from the original equation however that 2. $t^2+t^3+t^4... = 4t-t = 3t$.
Plug the value for $t^2+t^3+t^4...$ back into equation 1. That gives$ t= 3-3t$which gives$t=3/4$.
Let's check our answer.$S = a/(1-r) = 3/4(1-3/4) = 3$, which is indeed $4t$. So our answer is $t=3/4$.
Edit: got beaten to the answer :)
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