## kryton1212 2 years ago 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=\$????

1. kryton1212

(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.

2. kryton1212

@hartnn @ganeshie8 @primeralph

3. kryton1212

@Callisto

4. kryton1212

@mayankdevnani

5. xlegendx

i suppose not being in the tags means i still lack in notoriety huh

6. kryton1212

what?

7. kryton1212

nope, they help me a lot

8. kryton1212

@satellite73

9. hartnn

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...

10. hartnn

for b) part you will apply the sum formula instead of n'th term formula, with all other values being same

11. hartnn

you have those sum formulas ?

12. hartnn

and did you get that long explanation ? :P

13. kryton1212

why a1=1?

14. hartnn

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 \(S_n = a_1 [\dfrac{r^n-1}{r-1}]\)

15. hartnn

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

16. hartnn

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 :)

17. hartnn

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 =.... ?

18. kryton1212

oh, it seems easy. Let me try

19. kryton1212

but for part a. 100*(1* 0.2 ^(8) ) ???

20. hartnn

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 ?

21. kryton1212

uhh.

22. kryton1212

0.000256???!

23. hartnn

OMG, there must be some mistake in my method....it can't be 0.0256 let me review it...

24. hartnn

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 =.... ?

25. kryton1212

430?

26. hartnn

yes, i get 430 too :)

27. kryton1212

thanks. and part b Emily?

28. hartnn

revised formula for sum \(S_n = a_1 [\dfrac{(r+1)^n-1}{r}]\)

29. hartnn

a1 = 100, r =0.2, n =9 you'll get the answer :)

30. kryton1212

i know why i cannot get the answer. i think n=8... thankd

31. kryton1212

thanks*

32. hartnn

after 1 year means n=2, (100*0.2 is the 2nd term) so, after 8 years, means n=9

33. kryton1212

ah ha, i got it.

34. kryton1212

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?

35. hartnn

not total amount, IN THAT YEAR

36. kryton1212

okay..

37. hartnn

so, you will use n'th term formula or sum formula ?

38. kryton1212

n'th term

39. hartnn

correct! and since its \(\large 10th\) year, \(\large n=11\)

40. kryton1212

Emily=\$619.174 Bobby=\$600 then Emily's claim is correct, right?

41. hartnn

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 :)

42. kryton1212

and part c ii ? Bobby=\$7700? Emily=\$??

43. hartnn

ohh...for part 2 we need sum formula.....

44. kryton1212

yes

45. hartnn

Emily : 100 * [(1.2)^11 -1] / 0.2 =... ? i plugged in, r=0.2,a1 = 100, n=11

46. kryton1212

3215...

47. hartnn

48. kryton1212

49. hartnn

i am getting 3850 for bobby

50. hartnn

3850-3215=635

51. hartnn

Sum for bobby \(\large (11/2)[2*100+10*50]=3850\)

52. kryton1212

why 2 times 100?

53. hartnn

the formula \(\large (n/2) [\huge {\color{red} 2}a_1+(n-1)d]\)

54. kryton1212

uh? why? it's fixed

55. kryton1212

?

56. hartnn

yeah, thats the general formula for sum...

57. kryton1212

uh, OMG ..... okay, let me see. how about Emily's?

58. hartnn

sum formula i posted above, the "revised" one Emily : 100 * [(1.2)^11 -1] / 0.2

59. hartnn

\(S_n = a_1 [\dfrac{(r+1)^n-1}{r}]\)

60. kryton1212

ah ha, thanks

61. kryton1212

3215... thank you so much

62. hartnn

welcome ^_^