Please help... Only 2 days left for me on a summer school class and I am lost...
1. What is the sum of the geometric sequence 8, -16, 32 . if there are 15 terms? (1 point)
2. What is the sum of the geometric sequence 4, 12, 36 . if there are 9 terms? (1 point)
3. What is the sum of a 6-term geometric sequence if the first term is 11, the last term is -11,264 and the common ratio is -4? (1 point)
4. What is the sum of an 8-term geometric sequence if the first term is 10 and the last term is
781,250? (1 point)
Show all work as well

- anonymous

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- cwrw238

the formula for sum of n terms is
Sn = a1 * (1 - r^n)
------
1 - r

- theEric

http://www.mathsisfun.com/algebra/sequences-sums-geometric.html

- anonymous

But what do I plug in there to get the answers?

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## More answers

- theEric

The link I posted will walk you through what geometric sequences are, and that formula to get the answers!

- cwrw238

r = common ratio = second term / first term
a1 = first term
and n = number of terms

- cwrw238

so for question 1
r = -16/8 = -2
a1 = 8
and n = 15

- anonymous

So would the answer for the first one be 87384?

- theEric

@kathert I agree with 87384 :)

- cwrw238

8 * ( 1 - (-2)^15)
------------- = 87384
1 - (-2)

- anonymous

Oh yay! I got it right! Sorry I suck at math...

- cwrw238

A GS can be written as a, ar , ar^2 , ar^3 .......

- theEric

I've \(always\) been slow at math. But it took practice, and getting help, and I can do more math things now! Best of luck in your class! :) And it looks like you're in good hands with cwrw238 .

- anonymous

@theEric Thanks :)

- cwrw238

for the last probem you can find the common ratio r by dividing the 8th term by the first then you take the 7th root
8th term = ar^7
8th term / first term = ar^7 / a = r^7
781,250 / 10 = 78125
now use your calculator to find the 7th root of 78125
then use the sum formula
gotta go now

- anonymous

Whoa how do I do 7 root in my calculator??

- theEric

Do it like this: r ^ (- 7)

- theEric

No!

- theEric

Bad me!

- theEric

r ^ (1/7)

- theEric

That's what you want, \[\Huge r^{\frac{1}{7}}\]

- anonymous

Haha okay gimme a minute to figure this out...

- anonymous

I got 39364 for number 2 is that right?

- anonymous

Okay okay how do i do the last two??

- theEric

That's what I got for #2 as well.

- theEric

3. What is the sum of a 6-term geometric sequence if the first term is 11, the last term is -11,264 and the common ratio is -4? (1 point)
Well, the first term is your \(a\).\[a=11\]The last term is your \(a\ r^n\).\[a\ r^n=-11,264\]The common ratio is your \(r\).\[r=-4\]
You need \(\Large a\frac{1-r^n}{1-r}\).

- theEric

So, you have \(a\) and \(r\), and you need \(n\) or \(r^n\).

- theEric

Do you see how to get that?

- anonymous

uhhh.... no not really...so it would be -11 (1-(-4^n)/(1--4)?

- theEric

Yep! Hey, you know \(a\ r^n=-11,264\), so you can solve for \(r^n\)! That's how you'll finish that problem.

- theEric

General guideline: if you want something, solve for it.

- anonymous

wait so n is -11,246??

- theEric

No, \(a\ r^n=-11,264\). So you divide both sides by \(a=11\). That's algebra!

- anonymous

im sorry but im so lost.... where does the -11246 come in? do i set it equal to the equation?

- theEric

It is necessary to find the \(r^n\). Let me show you. Are you familiar with algebra?
\[a\ r^n = -11,264\]and\[a=11\]By substituting \(11\) for \(a\), which is okay because it's the same value either way, you'll get:\[11\ r^n=-11,264\]Now, you want to get \(r^n\) alone. So what you do is, you divide both sides by \(11\).
1. If the two sides are equal, and you do the same thing to both sides, both sides will still be equal!
2. Why divide by \(11\)? Well \(r^n\) is being multiplied by \(11\), and so you want to negate that. You want to make it be \(\Large \frac{\cancel{11}\ r^n}{\cancel{11}}\)
So, we left our equation off at\[11\ r^n=-11,264\]We divide by \(11\) to get\[\frac{11\ r^n}{11}=\frac{-11,264}{11}\]\[\frac{\cancel{11}\ r^n}{\cancel{11}}=\frac{-11,264}{11}\]\[r^n=\frac{-11,264}{11}\]

- theEric

Since you now have \(r^n\), \(a\), and \(r\), you can use that formula that you used for #1 and #2.

- anonymous

Ohhhhh okay! I get it now!

- theEric

SWEET! :) So, we'll both calculate #3 and see what we get....

- anonymous

I got -11 1/5

- theEric

I got \(2255\)...
Let me use Wolfram Alpha to double check. Then I can show you a link to the math.

- anonymous

okay

- theEric

http://www.wolframalpha.com/input/?i=11*%281-%28-11264%2F11%29%29%2F%281-%28-4%29%29

- theEric

Maybe you just had some calculator error.

- anonymous

Oh I see what I did wrong

- anonymous

How would I go about starting the last one?

- theEric

Well, I'm sure you know the formula you need to use, by now!\[\text{sum}=a\frac{(1-r^n)}{(1-r)}\]
4. What is the sum of an 8-term geometric sequence if the first term is 10 and the last term is
781,250? (1 point)
You need \(a\), \(r\), and \(r^n\), or \(n\).
What do you know from the problem, about the geometric sequence?

- theEric

Refresher:
\(a\) is the first term, or the common multiplier.
\(r\) is the common ratio.
\(n\) is the number of terms in the sequence.

- theEric

8-term \(\rightarrow n=8\)
first term is 10 \(\rightarrow a=10\)
last term is 781,250 \(\rightarrow a\ r^{n-1} =781,250\)

- anonymous

so would it be 10 (1-(781250/10))/1-r?

- anonymous

@theEric

- theEric

Sorry! Hi!

- anonymous

Its alright I disappeared for dinner so... haha

- theEric

Nope, sorry! Small mistake!
\[a\ r^{n-1} =781,250\]\(\qquad\Downarrow\qquad\)Substitute \(8\) for \(n\)
\[a\ r^{8-1} =a\ r^{7}=781,250\]\(\qquad\Downarrow\qquad\)Divide both sides by \(a\), and then substitute \(10\) in for \(a\)
\[r^7=\frac{781,250}{a}=\frac{781,250}{10}=78,125\]\(\qquad\Downarrow\qquad\)Get the seventh root of both sides\[\sqrt[7]{r^7}=r=\sqrt[7]{78,125}=5\]

- theEric

Well, you knew you didn't know \(r\), so I guess your only mistake was substituting \(r^n\) with \(r^{n-1}=781,250\), but you definitely had the right idea otherwise! I just found \(r\) for you, then...
Any questions on that part?
Now you have \(a\), \(n\), and \(r\).

- theEric

Which you can rearrange to spell \(r\ a\ n\): fun fact..

- anonymous

so it would be 8 *(1-78125)/(1-5) just to be sure
nice fun fact btw haha

- theEric

Haha, thanks! And check your "\(r^n\)", or \(5^8\).

- anonymous

so 390625?

- theEric

You got \(78125\) from \(r^{n-1}\), so I see where that came from :)
And, yep! \(390625\).

- anonymous

so that goes where i put the 78125

- theEric

Yep!

- theEric

It is \(r^n\), after all.

- theEric

and your formula is\[\text{sum}=a\frac{(1-r^n)}{(1-r)}\]

- anonymous

I got 781,248 for my answer

- theEric

I got the same! :) Congrats!

- theEric

Any questions about this problem?

- anonymous

Nope I'm good! Thank you for your help!

- theEric

You're welcome! Take care!

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