Diana lost 35 points on a project for school. She can make changes and receive three-tenths of the missing points back. She can do that 10 times. Create the formula for the sum of this geometric series and explain your steps in solving for the maximum number of points Diana can earn back.

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Diana lost 35 points on a project for school. She can make changes and receive three-tenths of the missing points back. She can do that 10 times. Create the formula for the sum of this geometric series and explain your steps in solving for the maximum number of points Diana can earn back.

Mathematics
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is there any way you could type that haha its very hard to interpret this. but thank you @triciaal
i'll type it :=) \[\huge\rm S_n =a+ar^{n-1}\] \[\large \rm (\frac{ 3 }{ 10 })^9 =a(1+r^{n-1})\] \[\large \rm 35 =\frac{ a(1+(\frac{ 3 }{ 10 })^9) }{ 1 -\frac{3}{10} }\] (formula: \[S_n = \frac{ a(1-r^n) }{ 1-r }\])

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So how would I plug this in?
Help ! haha please good person @ganeshie8
triciaal seems to be having the right method - i think you just need to add up the first 10 terms
remember the "partial sum formula" for geometric series ?
yes i do
\[\large S_n=a_1*\dfrac{1-r^n}{1-r}\] does it look something like above in ur notes ?
yes. but there is the sigma thing in the front
\[\large \sum\limits_{i=1}^na_i=a_1*\dfrac{1-r^n}{1-r}\] like that ?
yes
ok one sec
No wait, there is a mistake
ok
scratch that, il type it again fixing the mistake
ok thanks
Okay, good. Lets see whats going on in the given word problem. She can make changes and receive three-tenths of the missing points back, so this becomes the common ratio : \(r=\dfrac{3}{10}\) To start with, the missing points were 35, so the first term must be : \(a_1 = 35*\dfrac{3}{10} = 10.5\) Plug them in the partial sum formula and evaluate
\[\large \sum\limits_{i=1}^{10}a_i=10.5*\dfrac{1-\left(\frac{3}{10}\right)^{10}}{1-\frac{3}{10}}\] simplify..
yes thank you
let me know what you get and please medal triciaal, not me
I got 149
Wow! 149 ? She cannot receive 149 points back! double check ur work
oh sorry put it in way wrong haha
let me do it over
take your time..
wait yea i got 149 again lol
lets check with wolfram
ok
http://www.wolframalpha.com/input/?i=10.5*%281-%283%2F10%29%5E10%29%2F%281-3%2F10%29
ahh ok
wolf says\(14.9999\), which approximates to \(15\) looks you have just missed one decimal point somewhere..
but doesn't the sigma change it to 149
why ?
what do you know about sigma
like when u account for sigma in the problem.
what do you mean by that, could you elaborate
yea you have to take in to account the numbers above the sigma
idk maybe I'm crazy haha
\(\sum\) is just a short cut for "sum of terms" don't let that confuse you here
check this quick http://www.mathsisfun.com/algebra/sigma-notation.html
ohh thanks so much
so how many points is she gonna get back after 10 attempts ? whats the maximum possible points that she can get back if she is allowed to attempt infinitely many times ?
15 points !
careful here, \(14.999\) is not exactly same as \(15\) In fact, she can never get back \(15\) points even if she is allowed to attempt 100 or 1000 or any other big finite number of times
ah. so it would just be 14.999
thanks again
She gets back 14.999 after 10 attempts The maximum points that she can get back is 15 if she is allowed to try "infinite" times (which we know is practically impossible here)
http://www.wolframalpha.com/input/?i=Table%5B10.5*%281-%283%2F10%29%5En%29%2F%281-3%2F10%29%2C+%7Bn%2C+1%2C15%7D%5D
Notice that with each attempt, the sum is by increasing a tiny bit and getting closer to 15, but it never quite reaches 15. We say the sum "converges" to 15.

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