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kathert
 2 years ago
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 6term 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 8term geometric sequence if the first term is 10 and the last term is
781,250? (1 point)
Show all work as well
kathert
 2 years ago
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 6term 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 8term geometric sequence if the first term is 10 and the last term is 781,250? (1 point) Show all work as well

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cwrw238
 2 years ago
Best ResponseYou've already chosen the best response.1the formula for sum of n terms is Sn = a1 * (1  r^n)  1  r

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1http://www.mathsisfun.com/algebra/sequencessumsgeometric.html

kathert
 2 years ago
Best ResponseYou've already chosen the best response.0But what do I plug in there to get the answers?

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1The link I posted will walk you through what geometric sequences are, and that formula to get the answers!

cwrw238
 2 years ago
Best ResponseYou've already chosen the best response.1r = common ratio = second term / first term a1 = first term and n = number of terms

cwrw238
 2 years ago
Best ResponseYou've already chosen the best response.1so for question 1 r = 16/8 = 2 a1 = 8 and n = 15

kathert
 2 years ago
Best ResponseYou've already chosen the best response.0So would the answer for the first one be 87384?

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1@kathert I agree with 87384 :)

cwrw238
 2 years ago
Best ResponseYou've already chosen the best response.18 * ( 1  (2)^15)  = 87384 1  (2)

kathert
 2 years ago
Best ResponseYou've already chosen the best response.0Oh yay! I got it right! Sorry I suck at math...

cwrw238
 2 years ago
Best ResponseYou've already chosen the best response.1A GS can be written as a, ar , ar^2 , ar^3 .......

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1I'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 .

cwrw238
 2 years ago
Best ResponseYou've already chosen the best response.1for 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

kathert
 2 years ago
Best ResponseYou've already chosen the best response.0Whoa how do I do 7 root in my calculator??

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Do it like this: r ^ ( 7)

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1That's what you want, \[\Huge r^{\frac{1}{7}}\]

kathert
 2 years ago
Best ResponseYou've already chosen the best response.0Haha okay gimme a minute to figure this out...

kathert
 2 years ago
Best ResponseYou've already chosen the best response.0I got 39364 for number 2 is that right?

kathert
 2 years ago
Best ResponseYou've already chosen the best response.0Okay okay how do i do the last two??

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1That's what I got for #2 as well.

theEric
 2 years ago
Best ResponseYou've already chosen the best response.13. What is the sum of a 6term 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 \(\Large a\frac{1r^n}{1r}\).

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1So, you have \(a\) and \(r\), and you need \(n\) or \(r^n\).

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Do you see how to get that?

kathert
 2 years ago
Best ResponseYou've already chosen the best response.0uhhh.... no not really...so it would be 11 (1(4^n)/(14)?

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Yep! Hey, you know \(a\ r^n=11,264\), so you can solve for \(r^n\)! That's how you'll finish that problem.

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1General guideline: if you want something, solve for it.

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1No, \(a\ r^n=11,264\). So you divide both sides by \(a=11\). That's algebra!

kathert
 2 years ago
Best ResponseYou've already chosen the best response.0im sorry but im so lost.... where does the 11246 come in? do i set it equal to the equation?

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1It 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
 2 years ago
Best ResponseYou've already chosen the best response.1Since you now have \(r^n\), \(a\), and \(r\), you can use that formula that you used for #1 and #2.

kathert
 2 years ago
Best ResponseYou've already chosen the best response.0Ohhhhh okay! I get it now!

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1SWEET! :) So, we'll both calculate #3 and see what we get....

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1I got \(2255\)... Let me use Wolfram Alpha to double check. Then I can show you a link to the math.

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=11*%281%2811264%2F11%29%29%2F%281%284%29%29

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Maybe you just had some calculator error.

kathert
 2 years ago
Best ResponseYou've already chosen the best response.0Oh I see what I did wrong

kathert
 2 years ago
Best ResponseYou've already chosen the best response.0How would I go about starting the last one?

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Well, I'm sure you know the formula you need to use, by now!\[\text{sum}=a\frac{(1r^n)}{(1r)}\] 4. What is the sum of an 8term 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
 2 years ago
Best ResponseYou've already chosen the best response.1Refresher: \(a\) is the first term, or the common multiplier. \(r\) is the common ratio. \(n\) is the number of terms in the sequence.

theEric
 2 years ago
Best ResponseYou've already chosen the best response.18term \(\rightarrow n=8\) first term is 10 \(\rightarrow a=10\) last term is 781,250 \(\rightarrow a\ r^{n1} =781,250\)

kathert
 2 years ago
Best ResponseYou've already chosen the best response.0so would it be 10 (1(781250/10))/1r?

kathert
 2 years ago
Best ResponseYou've already chosen the best response.0Its alright I disappeared for dinner so... haha

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Nope, sorry! Small mistake! \[a\ r^{n1} =781,250\]\(\qquad\Downarrow\qquad\)Substitute \(8\) for \(n\) \[a\ r^{81} =a\ r^{7}=781,250\]\(\qquad\Downarrow\qquad\)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
 2 years ago
Best ResponseYou've already chosen the best response.1Well, you knew you didn't know \(r\), so I guess your only mistake was substituting \(r^n\) with \(r^{n1}=781,250\), 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
 2 years ago
Best ResponseYou've already chosen the best response.1Which you can rearrange to spell \(r\ a\ n\): fun fact..

kathert
 2 years ago
Best ResponseYou've already chosen the best response.0so it would be 8 *(178125)/(15) just to be sure nice fun fact btw haha

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Haha, thanks! And check your "\(r^n\)", or \(5^8\).

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1You got \(78125\) from \(r^{n1}\), so I see where that came from :) And, yep! \(390625\).

kathert
 2 years ago
Best ResponseYou've already chosen the best response.0so that goes where i put the 78125

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1It is \(r^n\), after all.

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1and your formula is\[\text{sum}=a\frac{(1r^n)}{(1r)}\]

kathert
 2 years ago
Best ResponseYou've already chosen the best response.0I got 781,248 for my answer

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1I got the same! :) Congrats!

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Any questions about this problem?

kathert
 2 years ago
Best ResponseYou've already chosen the best response.0Nope I'm good! Thank you for your help!

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1You're welcome! Take care!
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