anonymous
  • anonymous
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=$2700; Emily=$????
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
(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.
anonymous
  • anonymous
anonymous
  • anonymous

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anonymous
  • anonymous
anonymous
  • anonymous
i suppose not being in the tags means i still lack in notoriety huh
anonymous
  • anonymous
what?
anonymous
  • anonymous
nope, they help me a lot
anonymous
  • anonymous
hartnn
  • 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...
hartnn
  • hartnn
for b) part you will apply the sum formula instead of n'th term formula, with all other values being same
hartnn
  • hartnn
you have those sum formulas ?
hartnn
  • hartnn
and did you get that long explanation ? :P
anonymous
  • anonymous
why a1=1?
hartnn
  • 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 \(S_n = a_1 [\dfrac{r^n-1}{r-1}]\)
hartnn
  • 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
hartnn
  • 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 :)
hartnn
  • 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 =.... ?
anonymous
  • anonymous
oh, it seems easy. Let me try
anonymous
  • anonymous
but for part a. 100*(1* 0.2 ^(8) ) ???
hartnn
  • 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 ?
anonymous
  • anonymous
uhh.
anonymous
  • anonymous
0.000256???!
hartnn
  • hartnn
OMG, there must be some mistake in my method....it can't be 0.0256 let me review it...
hartnn
  • 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 =.... ?
anonymous
  • anonymous
430?
hartnn
  • hartnn
yes, i get 430 too :)
anonymous
  • anonymous
thanks. and part b Emily?
hartnn
  • hartnn
revised formula for sum \(S_n = a_1 [\dfrac{(r+1)^n-1}{r}]\)
hartnn
  • hartnn
a1 = 100, r =0.2, n =9 you'll get the answer :)
anonymous
  • anonymous
i know why i cannot get the answer. i think n=8... thankd
anonymous
  • anonymous
thanks*
hartnn
  • hartnn
after 1 year means n=2, (100*0.2 is the 2nd term) so, after 8 years, means n=9
anonymous
  • anonymous
ah ha, i got it.
anonymous
  • anonymous
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?
hartnn
  • hartnn
not total amount, IN THAT YEAR
anonymous
  • anonymous
okay..
hartnn
  • hartnn
so, you will use n'th term formula or sum formula ?
anonymous
  • anonymous
n'th term
hartnn
  • hartnn
correct! and since its \(\large 10th\) year, \(\large n=11\)
anonymous
  • anonymous
Emily=$619.174 Bobby=$600 then Emily's claim is correct, right?
hartnn
  • 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 :)
anonymous
  • anonymous
and part c ii ? Bobby=$7700? Emily=$??
hartnn
  • hartnn
ohh...for part 2 we need sum formula.....
anonymous
  • anonymous
yes
hartnn
  • hartnn
Emily : 100 * [(1.2)^11 -1] / 0.2 =... ? i plugged in, r=0.2,a1 = 100, n=11
anonymous
  • anonymous
3215...
hartnn
  • hartnn
is that the answer in your book ? difference = 7700-3215
anonymous
  • anonymous
the answer is 635
hartnn
  • hartnn
i am getting 3850 for bobby
hartnn
  • hartnn
3850-3215=635
hartnn
  • hartnn
Sum for bobby \(\large (11/2)[2*100+10*50]=3850\)
anonymous
  • anonymous
why 2 times 100?
hartnn
  • hartnn
the formula \(\large (n/2) [\huge {\color{red} 2}a_1+(n-1)d]\)
anonymous
  • anonymous
uh? why? it's fixed
anonymous
  • anonymous
?
hartnn
  • hartnn
yeah, thats the general formula for sum...
anonymous
  • anonymous
uh, OMG ..... okay, let me see. how about Emily's?
hartnn
  • hartnn
sum formula i posted above, the "revised" one Emily : 100 * [(1.2)^11 -1] / 0.2
hartnn
  • hartnn
\(S_n = a_1 [\dfrac{(r+1)^n-1}{r}]\)
anonymous
  • anonymous
ah ha, thanks
anonymous
  • anonymous
3215... thank you so much
hartnn
  • hartnn
welcome ^_^

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