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kathert
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
the formula for sum of n terms is Sn = a1 * (1 - r^n) ------ 1 - r
http://www.mathsisfun.com/algebra/sequences-sums-geometric.html
But what do I plug in there to get the answers?
The link I posted will walk you through what geometric sequences are, and that formula to get the answers!
r = common ratio = second term / first term a1 = first term and n = number of terms
so for question 1 r = -16/8 = -2 a1 = 8 and n = 15
So would the answer for the first one be 87384?
@kathert I agree with 87384 :)
8 * ( 1 - (-2)^15) ------------- = 87384 1 - (-2)
Oh yay! I got it right! Sorry I suck at math...
A GS can be written as a, ar , ar^2 , ar^3 .......
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 .
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
Whoa how do I do 7 root in my calculator??
Do it like this: r ^ (- 7)
That's what you want, \[\Huge r^{\frac{1}{7}}\]
Haha okay gimme a minute to figure this out...
I got 39364 for number 2 is that right?
Okay okay how do i do the last two??
That's what I got for #2 as well.
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 \(\Large a\frac{1-r^n}{1-r}\).
So, you have \(a\) and \(r\), and you need \(n\) or \(r^n\).
Do you see how to get that?
uhhh.... no not really...so it would be -11 (1-(-4^n)/(1--4)?
Yep! Hey, you know \(a\ r^n=-11,264\), so you can solve for \(r^n\)! That's how you'll finish that problem.
General guideline: if you want something, solve for it.
No, \(a\ r^n=-11,264\). So you divide both sides by \(a=11\). That's algebra!
im sorry but im so lost.... where does the -11246 come in? do i set it equal to the equation?
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\[\frac{11\ r^n}{11}=\frac{-11,264}{11}\]\[\frac{\cancel{11}\ r^n}{\cancel{11}}=\frac{-11,264}{11}\]\[r^n=\frac{-11,264}{11}\]
Since you now have \(r^n\), \(a\), and \(r\), you can use that formula that you used for #1 and #2.
Ohhhhh okay! I get it now!
SWEET! :) So, we'll both calculate #3 and see what we get....
I got \(2255\)... Let me use Wolfram Alpha to double check. Then I can show you a link to the math.
http://www.wolframalpha.com/input/?i=11*%281-%28-11264%2F11%29%29%2F%281-%28-4%29%29
Maybe you just had some calculator error.
Oh I see what I did wrong
How would I go about starting the last one?
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?
Refresher: \(a\) is the first term, or the common multiplier. \(r\) is the common ratio. \(n\) is the number of terms in the sequence.
8-term \(\rightarrow n=8\) first term is 10 \(\rightarrow a=10\) last term is 781,250 \(\rightarrow a\ r^{n-1} =781,250\)
so would it be 10 (1-(781250/10))/1-r?
Its alright I disappeared for dinner so... haha
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\[\sqrt[7]{r^7}=r=\sqrt[7]{78,125}=5\]
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 \(r\).
Which you can rearrange to spell \(r\ a\ n\): fun fact..
so it would be 8 *(1-78125)/(1-5) just to be sure nice fun fact btw haha
Haha, thanks! And check your "\(r^n\)", or \(5^8\).
You got \(78125\) from \(r^{n-1}\), so I see where that came from :) And, yep! \(390625\).
so that goes where i put the 78125
It is \(r^n\), after all.
and your formula is\[\text{sum}=a\frac{(1-r^n)}{(1-r)}\]
I got 781,248 for my answer
I got the same! :) Congrats!
Any questions about this problem?
Nope I'm good! Thank you for your help!
You're welcome! Take care!