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kryton1212 Group Title

Bobby and Emily each got $100 red packet money from their grandmother when they were born. Then, their grandmother gave them red packet money on each of their birthdays. The amount of money given to Bobby was $50 more than the previous year, while the amount given to Emily was 20% more than the previous year. (a) How much did each of them get when they were 8 years old? --->Bobby=$500; Emily=$???? (b) Find the total amount of red packet money that each of them got until they were 8 years old. --->Bobby=$2700; Emily=$????

  • one year ago
  • one year ago

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  1. kryton1212 Group Title
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    (c) Emily claims that she will get more packet money than Bobby when they are both 10 years old. (i) is Emily's claim correct? Why? (ii) Find the difference between their total red packet money when they are 10 years old.

    • one year ago
  2. kryton1212 Group Title
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    @hartnn @ganeshie8 @primeralph

    • one year ago
  3. kryton1212 Group Title
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    @Callisto

    • one year ago
  4. kryton1212 Group Title
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    @mayankdevnani

    • one year ago
  5. xlegendx Group Title
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    i suppose not being in the tags means i still lack in notoriety huh

    • one year ago
  6. kryton1212 Group Title
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    what?

    • one year ago
  7. kryton1212 Group Title
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    nope, they help me a lot

    • one year ago
  8. kryton1212 Group Title
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    @satellite73

    • one year ago
  9. hartnn Group Title
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    ok, the first part is just an arithmetic sequence, for bobby, (100,150,200,250....) so, first term = a1 = 100 and common difference = 50 9th term (after 8 years) = a1 + (n-1)d = 100 + 8*50 = $500 (correct) but for emily : (100 + 100*0.2 +100*0.2*0.2+...) = 100 (1+0.2 +0.2^2+0.2^3+...) now this is geometric sequence a1 = 1, r = 0.2 so, for 9th term (after 8 years) = a1 (r)^(n-1) = 1 * (0.2)^8 =.... ? multiply the result by 100 to get the answer...

    • one year ago
  10. hartnn Group Title
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    for b) part you will apply the sum formula instead of n'th term formula, with all other values being same

    • one year ago
  11. hartnn Group Title
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    you have those sum formulas ?

    • one year ago
  12. hartnn Group Title
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    and did you get that long explanation ? :P

    • one year ago
  13. kryton1212 Group Title
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    why a1=1?

    • one year ago
  14. hartnn Group Title
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    for bobby, Arithmetic sequence n'th term formula \(T_n = a_1+(n-1)d\) sum formula \(S_n = (n/2)[2a_1+(n-1)d]\) for emily , geometric sequence : n'th term formula \(T_n = a_1r^{n-1}\) sum formula \(S_n = a_1 [\dfrac{r^n-1}{r-1}]\)

    • one year ago
  15. hartnn Group Title
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    for emily, the sequence will be 100 , 100*0.2 , 100*0.2*0.2 , .... any doubts in this ? so, taking 100 out, 100 ( 1, 0.2, 0.2*0.2,....) in the above sequence first term = a1 = 1

    • one year ago
  16. hartnn Group Title
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    if you want you can consider the sequence as 100, 100*0.2 , 100*0.2*0.2,.... then a1 = 1st term = 100 r= 0.2 (same) this will be easier :)

    • one year ago
  17. hartnn Group Title
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    but for emily : (100 + 100*0.2 +100*0.2*0.2+...) now this is geometric sequence a1 = 100, r = 0.2 so, for 9th term (after 8 years) = a1 (r)^(n-1) = 100 * (0.2)^8 =.... ?

    • one year ago
  18. kryton1212 Group Title
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    oh, it seems easy. Let me try

    • one year ago
  19. kryton1212 Group Title
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    but for part a. 100*(1* 0.2 ^(8) ) ???

    • one year ago
  20. hartnn Group Title
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    a1= 100, r = 0.2 , n= 9 \(\large T_9 = a_1r^{n-1} = 100(0.2)^8 =...?\) use calculator.....what exactly wast he doubt ?

    • one year ago
  21. kryton1212 Group Title
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    uhh.

    • one year ago
  22. kryton1212 Group Title
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    0.000256???!

    • one year ago
  23. hartnn Group Title
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    OMG, there must be some mistake in my method....it can't be 0.0256 let me review it...

    • one year ago
  24. hartnn Group Title
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    ok, its like compound interest formula! let me revise the formula n'th term \(T_n = a_1 (1+r)^{n-1}\) here, a1 = 100, r=0.2, n = 9 gives T9 = 100 (1.2)^8 =.... ?

    • one year ago
  25. kryton1212 Group Title
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    430?

    • one year ago
  26. hartnn Group Title
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    yes, i get 430 too :)

    • one year ago
  27. kryton1212 Group Title
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    thanks. and part b Emily?

    • one year ago
  28. hartnn Group Title
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    revised formula for sum \(S_n = a_1 [\dfrac{(r+1)^n-1}{r}]\)

    • one year ago
  29. hartnn Group Title
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    a1 = 100, r =0.2, n =9 you'll get the answer :)

    • one year ago
  30. kryton1212 Group Title
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    i know why i cannot get the answer. i think n=8... thankd

    • one year ago
  31. kryton1212 Group Title
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    thanks*

    • one year ago
  32. hartnn Group Title
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    after 1 year means n=2, (100*0.2 is the 2nd term) so, after 8 years, means n=9

    • one year ago
  33. kryton1212 Group Title
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    ah ha, i got it.

    • one year ago
  34. kryton1212 Group Title
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    i want to know , in part c, the statement said "Emily claims that she will get more red packet money than Bobby when they are both 10 years old", total amount or the money in that year?

    • one year ago
  35. hartnn Group Title
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    not total amount, IN THAT YEAR

    • one year ago
  36. kryton1212 Group Title
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    okay..

    • one year ago
  37. hartnn Group Title
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    so, you will use n'th term formula or sum formula ?

    • one year ago
  38. kryton1212 Group Title
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    n'th term

    • one year ago
  39. hartnn Group Title
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    correct! and since its \(\large 10th\) year, \(\large n=11\)

    • one year ago
  40. kryton1212 Group Title
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    Emily=$619.174 Bobby=$600 then Emily's claim is correct, right?

    • one year ago
  41. hartnn Group Title
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    for bobby first term = a1 = 100 and common difference = 50 11th term (after 10 years) = a1 + (n-1)d = 100 + 10*50 =600 (correct ) T11 = 100 * (1.2)^10 = 619.174 (correct) and YES , emily's claim was correct :)

    • one year ago
  42. kryton1212 Group Title
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    and part c ii ? Bobby=$7700? Emily=$??

    • one year ago
  43. hartnn Group Title
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    ohh...for part 2 we need sum formula.....

    • one year ago
  44. kryton1212 Group Title
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    yes

    • one year ago
  45. hartnn Group Title
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    Emily : 100 * [(1.2)^11 -1] / 0.2 =... ? i plugged in, r=0.2,a1 = 100, n=11

    • one year ago
  46. kryton1212 Group Title
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    3215...

    • one year ago
  47. hartnn Group Title
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    is that the answer in your book ? difference = 7700-3215

    • one year ago
  48. kryton1212 Group Title
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    the answer is 635

    • one year ago
  49. hartnn Group Title
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    i am getting 3850 for bobby

    • one year ago
  50. hartnn Group Title
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    3850-3215=635

    • one year ago
  51. hartnn Group Title
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    Sum for bobby \(\large (11/2)[2*100+10*50]=3850\)

    • one year ago
  52. kryton1212 Group Title
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    why 2 times 100?

    • one year ago
  53. hartnn Group Title
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    the formula \(\large (n/2) [\huge {\color{red} 2}a_1+(n-1)d]\)

    • one year ago
  54. kryton1212 Group Title
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    uh? why? it's fixed

    • one year ago
  55. kryton1212 Group Title
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    ?

    • one year ago
  56. hartnn Group Title
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    yeah, thats the general formula for sum...

    • one year ago
  57. kryton1212 Group Title
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    uh, OMG ..... okay, let me see. how about Emily's?

    • one year ago
  58. hartnn Group Title
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    sum formula i posted above, the "revised" one Emily : 100 * [(1.2)^11 -1] / 0.2

    • one year ago
  59. hartnn Group Title
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    \(S_n = a_1 [\dfrac{(r+1)^n-1}{r}]\)

    • one year ago
  60. kryton1212 Group Title
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    ah ha, thanks

    • one year ago
  61. kryton1212 Group Title
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    3215... thank you so much

    • one year ago
  62. hartnn Group Title
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    welcome ^_^

    • one year ago
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