MHB Geometric Sequences help - (3 given terms, find the rest)

[jaxx]

I need to find the value of the first term for this geometric series.

Sn = 33
tn = 48
r = -2

I know that I have to take the formulas tn = t1 x r^(n-1), and Sn = [t1 x (r^n) - 1] / (r - 1), and isolate t1 for the first formula and then input that into the second, but I don't know the actual process of doing that.

If anyone can explain it step by step and with reasons maybe, that would be really helpful, I'm not good at math and get confused easily. Any help is appreciated

 

[MarkFL]

We are given:

$$S_n=t_1\frac{1-(-2)^n}{1-(-2)}=t_1\frac{1-(-2)^n}{3}=33\,\therefore\,t_1=\frac{99}{1-(-2)^n}$$

$$t_n=t_1(-2)^{n-1}=48\,\therefore\,(-2)^n=-\frac{96}{t_1}$$

Now, substitute for $$(-2)^n$$ into the expression for $$t_1$$ to get:

$$t_1=\frac{99}{1+\frac{96}{t_1}}$$

Can you proceed?

 

[jaxx]

sorry I'm not sure how you got to t1 = 99 / 1 - (-2)^n.

I get to 33 = t1[1 - (-2)^n] / 3, then wouldn't you multiply by 3 to get rid of the denominator, making it 99 = t1[1 - (-2)^n]?

either way I see you did multiply by 3, but I'm not sure what you did to flip the equation to get 1 - (-2)^n as the denominator, and Sn as the numerator.also when substituting in 99 / 1 - (-2)^n as t1 in the other formula, I'm not sure how to solve that. The equation would be 48 = [99 / 1 - (-2)^n] x (-2)^(n-1) right?

From there I'm not really sure how to go about solving it, I don't know if it's too much work for you to explain it all or not, but thanks for responding anyway.

Also at the end when you get (-2)^n = 96/t1, and then end up with t1 = (-2)^n, why does that 96 disappear? Or am I reading it wrong?

 

[MarkFL]

sorry I'm not sure how you got to t1 = 99 / 1 - (-2)^n.

I get to 33 = t1[1 - (-2)^n] / 3, then wouldn't you multiply by 3 to get rid of the denominator, making it 99 = t1[1 - (-2)^n]?

either way I see you did multiply by 3, but I'm not sure what you did to flip the equation to get 1 - (-2)^n as the denominator, and Sn as the numerator.

Okay, you are clear on how we get to:

$$99=t_1\left(1-(-2)^n \right)$$

Now, we may divide both sides by $$1-(-2)^n$$ to get:

$$\frac{99}{1-(-2)^n}=\frac{t_1\left(1-(-2)^n \right)}{1-(-2)^n}$$

On the right, divide out or cancel like factors:

$$\frac{99}{1-(-2)^n}=\frac{t_1\cancel{\left(1-(-2)^n \right)}}{\cancel{1-(-2)^n}}$$

And we are left with:

$$\frac{99}{1-(-2)^n}=t_1$$

...also when substituting in 99 / 1 - (-2)^n as t1 in the other formula, I'm not sure how to solve that.

What I did was solve for $(-2)^n$ with the second equation, and then substituted that into the first so that $t_1$ is the only unknown, and this is the unknown we are trying to find.

...The equation would be 48 = [99 / 1 - (-2)^n] x (-2)^(n-1) right?

From there I'm not really sure how to go about solving it, I don't know if it's too much work for you to explain it all or not, but thanks for responding anyway.

This would allow you to find $n$, and then you could find $t_1$ from this, but I think my suggested method is more straightforward.

...Also at the end when you get (-2)^n = 96/t1, and then end up with t1 = (-2)^n, why does that 96 disappear? Or am I reading it wrong?

I actually get:

$$(-2)^n=-\frac{96}{t_1}$$

because I multiplied both sides of $$t_1(-2)^{n-1}=48$$ by $\frac{-2}{t_1}$.

I believe you are reading it wrong, as I substituted for $$(-2)^n=-\frac{96}{t_1}$$ to get:

$$t_1=\frac{99}{1-\left(-\frac{96}{t_1} \right)}$$

$$t_1=\frac{99}{1+\frac{96}{t_1}}$$

You see, the $96$ is still there. Now we have an equation where the only unknown is $t_1$, which is what we seek. As the next step, I suggest multiplying both sides by:

$$1+\frac{96}{t_1}$$

What do you get when you do this?

 

[jaxx]

Right, I get the first part now.

I get to t1 = -99 / -2^n - 1, and then substitute it into the tn formula like you did, which makes

48 = (-99 / -2^n - 1) [-2^(n-1)], and then multiply by -2 because it looks like that's what you did to get rid of it in the denominator.

So I get -96 = - 99 / (n - 1) x [-2^(n-1)], but from there I'm not sure what the next step is. I could multiply by n - 1 to get rid of the fraction, but that would just put the variable on both sides of the equation, which isn't what I want to do right?

If what I did was right, what's the next step?

 

[MarkFL]

It looks like you are still trying to solve for $n$ instead of $t_1$. If you follow my suggestion, you need not worry about finding $n$, as you get rid of it by substitution and have an equation in one variable, the one we are asked to find.

 

[jaxx]

I guess I just don't understand at all why you would multiply both sides of t1(−2)n−1=48 by -2 / t1, and much less how to actually work through it from there.

I get why your way is easier, and it finds t1 instead of taking the extra step to find n and then produce t1, but if possible, would you mind showing me how to solve from here?

-96 = - 99 / (n - 1) x [-2^(n-1)]

I understand it a lot more if I do it that way, even if it might be longer and takes the extra step to produce both variables instead of just the one I need.

Thanks for all your help so far as well.

 

[MarkFL]

No problem, I just wanted to point out that you get there more quickly the other way. So, let's go back to:

$$t_1\frac{1-(-2)^n}{3}=33$$

$$t_1(-2)^{n-1}=48$$

We may solving both for $t_1$ and then equate the results:

$$\frac{99}{1-(-2)^n}=\frac{48}{(-2)^{n-1}}$$

Multiplying the left side by $$-1=\frac{1}{-1}$$ and the right side by $$-1=\frac{2}{-2}$$ we get:

$$\frac{99}{(-2)^n-1}=\frac{96}{(-2)^{n}}$$

Cross-multiplying, we find:

$$99(-2)^n=96(-2)^n-96$$

Subtracting through by $$96(-2)^n$$, we find:

$$3(-2)^n=-96$$

Dividing through by $3$, we get:

$$(-2)^n=-32=(-2)^5$$

And so we must have:

$$n=5$$

Now you can find $$t_1$$ from this.

Using my suggestion above, we find:

$$t_1+96=99$$

$$t_1=3$$

This is the value you should find for $n=5$.

 

[jaxx]

So why did you multiply by -1 = 1/-1? Can you not just multiply by 1/-1? Not sure I understand that bit, and multiplying by 2 / -2 gets rid of the -1 in (n - 1)? Also don't you have to multiply both sides of an equation by the same thing? Even though it is essentially the same but the negatives are different, I thought it had to be the exact same.

Also I'm not sure what you mean by 'subtracting through', when you simplify

99(−2)^n = 96 (−2)^n − 96

to get 3(−2)^n= −96.

I'm not sure how to simplify that, mainly I'm not sure how to get rid of the 2 "n" powers, or at least consolidate them into one.

 

[soroban]

Hello, jaxx!

Another approach . . .

\text{Given: }\;\begin{Bmatrix}S_n &=& 33 \\ t_n &=& 48 \\ r &=& \text{-}2 \end{Bmatrix}

\text{Find the first term, }t_1

\text{Formula: }\:t_n \:=\:t_1r^{n-1}

\text{Then: }\:t_n \:=\:t_1(\text{-}2)^{n-1} \:=\:48

\text{Hence: }\:t_1(\text{-}2)^n \:=\:\text{-}96\;\;\color{blue}{[1]}\text{Formula: }\:S_n \:=\:t_1\frac{1-r^n}{1-r}

\text{Then: }\:S_n \:=\:t_1\frac{1-(\text{-}2)^n}{1-(\text{-}2)} \:=\:33

\text{Hence: }\:t_1\left[1-(\text{-}2)^n \right]\:=\:99\text{We have: }\:t_1 - t_1(\text{-}2)^n \:=\:99

\text{Substitute }\color{blue}{[1]}:\;t_1 - (\text{-}96) \:=\:99

\text{Therefore: }\:t_1 \:=\:3

 

[MarkFL]

So why did you multiply by -1 = 1/-1? Can you not just multiply by 1/-1? Not sure I understand that bit, and multiplying by 2 / -2 gets rid of the -1 in (n - 1)? Also don't you have to multiply both sides of an equation by the same thing? Even though it is essentially the same but the negatives are different, I thought it had to be the exact same.

Also I'm not sure what you mean by 'subtracting through', when you simplify

99(−2)^n = 96 (−2)^n − 96

to get 3(−2)^n= −96.

I'm not sure how to simplify that, mainly I'm not sure how to get rid of the 2 "n" powers, or at least consolidate them into one.

There are usually many ways to solve an equation, it really boils down to personal preference, as long as the operations are all legal.

When I say "subtracting through," this means subtracting from both sides.

Once you get to:

$$3(-2)^n=-96$$

dividing both sides by 3 gives:

$$(-2)^n=-32$$

At this point, one can observe that $$-32=(-2)^5$$ and then you are left to equate the exponents to get:

$n=5$.

 

[jaxx]

Hello, jaxx!

Another approach . . .


\text{Formula: }\:t_n \:=\:t_1r^{n-1}

\text{Then: }\:t_n \:=\:t_1(\text{-}2)^{n-1} \:=\:48

\text{Hence: }\:t_1(\text{-}2)^n \:=\:\text{-}96\;\;\color{blue}{[1]}\text{Formula: }\:S_n \:=\:t_1\frac{1-r^n}{1-r}

\text{Then: }\:S_n \:=\:t_1\frac{1-(\text{-}2)^n}{1-(\text{-}2)} \:=\:33

\text{Hence: }\:t_1\left[1-(\text{-}2)^n \right]\:=\:99\text{We have: }\:t_1 - t_1(\text{-}2)^n \:=\:99

\text{Substitute }\color{blue}{[1]}:\;t_1 - (\text{-}96) \:=\:99

\text{Therefore: }\:t_1 \:=\:3

Hence: t1[1−(-2)n]=99We have: t1−t1(-2)n=99why did it become t1 - t1? where did that come from?also, how did you just get rid of "n" at the end? there was never a determined = t1 equation, and when you substituted [1] that had n in it as well, so how did it disappear?

 

[jaxx]

There are usually many ways to solve an equation, it really boils down to personal preference, as long as the operations are all legal.

When I say "subtracting through," this means subtracting from both sides.

Once you get to:

$$3(-2)^n=-96$$

dividing both sides by 3 gives:

$$(-2)^n=-32$$

At this point, one can observe that $$-32=(-2)^5$$ and then you are left to equate the exponents to get:

$n=5$.

just out of curiosity, was I right when I got to this?

-96 = - 99 / (n - 1) x [-2^(n-1)]

if so, am I right by multiplying by -2 to get rid of the (^n - 1), and ending up with -192 = (99 / n - 1) (-2n), dividing by -2, getting 96 = (99 / n - 1) x n? I don't know what to do from there but was I right up until that point?also what I meant was that I wasn't sure how

99(−2)^n = 96 (−2)^n − 96

can become 3(−2)^n= −96. sorry for this taking so long, and thanks for the help

 

[MarkFL]

just out of curiosity, was I right when I got to this?

-96 = - 99 / (n - 1) x [-2^(n-1)]...

No, the denominator on the right is incorrect, it should be:

$$(-2)^{n}-1$$

Also, you should be aware that:

$$(-a)^n\ne -a^n$$

There is an order of operations issue at work. On the left, $-a$ is being raised to the $n$th power, while on the right $a$ is raised to the $n$th power, and then negated. If $n$ is odd, then they will be equal, but only because $(-1)^{2n+1}=-1$.

...
also what I meant was that I wasn't sure how

99(−2)^n = 96 (−2)^n − 96

can become 3(−2)^n= −96. sorry for this taking so long, and thanks for the help

Okay, we have:

$$99(-2)^n=96(-2)^n-96$$

Subtracting $$96(-2)^n$$ from both sides, we get:

$$99(-2)^n-96(-2)^n=96(-2)^n-96(-2)^n-96$$

Now, if this is not clear that we have $$3(-2)^n$$ on the left and $-96$ on the right, then we may factor as follows:

$$(-2)^n(99-96)=(-2)^n(96-96)-96$$

Now, combine like terms:

$$(-2)^n(3)=(-2)^n(0)-96$$

and rewrite:

$$3(-2)^n=-96$$

Does this make more sense?

And please do not worry about how long it takes...we wish for our members posting questions to fully understand how to solve the problem. :D

 

[jaxx]

No, the denominator on the right is incorrect, it should be:

$$(-2)^{n}-1$$

Also, you should be aware that:

$$(-a)^n\ne -a^n$$

There is an order of operations issue at work. On the left, $-a$ is being raised to the $n$th power, while on the right $a$ is raised to the $n$th power, and then negated. If $n$ is odd, then they will be equal, but only because $(-1)^{2n+1}=-1$.
Okay, we have:

$$99(-2)^n=96(-2)^n-96$$

Subtracting $$96(-2)^n$$ from both sides, we get:

$$99(-2)^n-96(-2)^n=96(-2)^n-96(-2)^n-96$$

Now, if this is not clear that we have $$3(-2)^n$$ on the left and $-96$ on the right, then we may factor as follows:

$$(-2)^n(99-96)=(-2)^n(96-96)-96$$

Now, combine like terms:

$$(-2)^n(3)=(-2)^n(0)-96$$

and rewrite:

$$3(-2)^n=-96$$

Does this make more sense?

And please do not worry about how long it takes...we wish for our members posting questions to fully understand how to solve the problem. :D

Alright thanks, my only question now is what it means to "combine like terms", and how you get 3 from 2^n(99 - 96)

 

[MarkFL]

The expressions:

$$99-96$$ and $$96-96$$

have like terms in that they contain numbers that may be subtracted to get:

$$99-96=3$$ and $$96-96=0$$

 

[jaxx]

The expressions:

$$99-96$$ and $$96-96$$

have like terms in that they contain numbers that may be subtracted to get:

$$99-96=3$$ and $$96-96=0$$

so for 99(-2)^n - 96 (-2)^n, since 3 factors into both 99 and 96, you can simply make it 3(-2n)?

 

[MarkFL]

so for 99(-2)^n - 96 (-2)^n, since 3 factors into both 99 and 96, you can simply make it 3(-2n)?

No, we have 99 of something, and we are taking away 96 of them, leaving 3. For example:

$$99x-96x=3x$$

$$99u^3-96u^3=3u^3$$

$$99\sin(t)-96\sin(t)=3\sin(t)$$

$$99\int f(x)\,dx-96\int f(x)\,dx=3\int f(x)\,dx$$

 

[jaxx]

No, we have 99 of something, and we are taking away 96 of them, leaving 3. For example:

$$99x-96x=3x$$

$$99u^3-96u^3=3u^3$$

$$99\sin(t)-96\sin(t)=3\sin(t)$$

$$99\int f(x)\,dx-96\int f(x)\,dx=3\int f(x)\,dx$$

oh it's just subtraction, I see. Thanks very much for all your help then, I really appreciate it.