## mmend98 one year ago 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.

1. triciaal

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2. mmend98

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

3. 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: $S_n = \frac{ a(1-r^n) }{ 1-r }$)

4. mmend98

So how would I plug this in?

5. mmend98

@triciaal

6. mmend98

Help ! haha please good person @ganeshie8

7. ganeshie8

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

8. ganeshie8

remember the "partial sum formula" for geometric series ?

9. mmend98

yes i do

10. ganeshie8

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

11. mmend98

yes. but there is the sigma thing in the front

12. ganeshie8

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

13. mmend98

yes

14. mmend98

ok one sec

15. ganeshie8

No wait, there is a mistake

16. mmend98

ok

17. ganeshie8

scratch that, il type it again fixing the mistake

18. mmend98

ok thanks

19. 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

20. ganeshie8

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

21. mmend98

yes thank you

22. ganeshie8

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

23. mmend98

I got 149

24. ganeshie8

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

25. mmend98

oh sorry put it in way wrong haha

26. mmend98

let me do it over

27. ganeshie8

28. mmend98

wait yea i got 149 again lol

29. ganeshie8

lets check with wolfram

30. mmend98

ok

31. ganeshie8
32. mmend98

ahh ok

33. ganeshie8

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

34. mmend98

but doesn't the sigma change it to 149

35. ganeshie8

why ?

36. ganeshie8

what do you know about sigma

37. mmend98

like when u account for sigma in the problem.

38. ganeshie8

what do you mean by that, could you elaborate

39. mmend98

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

40. mmend98

idk maybe I'm crazy haha

41. ganeshie8

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

42. ganeshie8

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

43. mmend98

ohh thanks so much

44. 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 ?

45. mmend98

15 points !

46. 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

47. mmend98

ah. so it would just be 14.999

48. mmend98

thanks again

49. 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)

50. ganeshie8
51. 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.