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.

- mmend98

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- triciaal

|dw:1439420130504:dw|

- mmend98

is there any way you could type that haha its very hard to interpret this. but thank you @triciaal

- Nnesha

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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## More answers

- mmend98

So how would I plug this in?

- mmend98

@triciaal

- mmend98

Help ! haha please good person @ganeshie8

- ganeshie8

triciaal seems to be having the right method - i think you just need to add up the first 10 terms

- ganeshie8

remember the "partial sum formula" for geometric series ?

- mmend98

yes i do

- ganeshie8

\[\large S_n=a_1*\dfrac{1-r^n}{1-r}\]
does it look something like above in ur notes ?

- mmend98

yes. but there is the sigma thing in the front

- ganeshie8

\[\large \sum\limits_{i=1}^na_i=a_1*\dfrac{1-r^n}{1-r}\]
like that ?

- mmend98

yes

- mmend98

ok one sec

- ganeshie8

No wait, there is a mistake

- mmend98

ok

- ganeshie8

scratch that, il type it again fixing the mistake

- mmend98

ok thanks

- ganeshie8

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

- ganeshie8

\[\large \sum\limits_{i=1}^{10}a_i=10.5*\dfrac{1-\left(\frac{3}{10}\right)^{10}}{1-\frac{3}{10}}\]
simplify..

- mmend98

yes thank you

- ganeshie8

let me know what you get and please medal triciaal, not me

- mmend98

I got 149

- ganeshie8

Wow! 149 ?
She cannot receive 149 points back! double check ur work

- mmend98

oh sorry put it in way wrong haha

- mmend98

let me do it over

- ganeshie8

take your time..

- mmend98

wait yea i got 149 again lol

- ganeshie8

lets check with wolfram

- mmend98

ok

- ganeshie8

http://www.wolframalpha.com/input/?i=10.5*%281-%283%2F10%29%5E10%29%2F%281-3%2F10%29

- mmend98

ahh ok

- ganeshie8

wolf says\(14.9999\), which approximates to \(15\)
looks you have just missed one decimal point somewhere..

- mmend98

but doesn't the sigma change it to 149

- ganeshie8

why ?

- ganeshie8

what do you know about sigma

- mmend98

like when u account for sigma in the problem.

- ganeshie8

what do you mean by that, could you elaborate

- mmend98

yea you have to take in to account the numbers above the sigma

- mmend98

idk maybe I'm crazy haha

- ganeshie8

\(\sum\) is just a short cut for "sum of terms"
don't let that confuse you here

- ganeshie8

check this quick
http://www.mathsisfun.com/algebra/sigma-notation.html

- mmend98

ohh thanks so much

- ganeshie8

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 ?

- mmend98

15 points !

- ganeshie8

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

- mmend98

ah. so it would just be 14.999

- mmend98

thanks again

- ganeshie8

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)

- ganeshie8

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

- ganeshie8

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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