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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
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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.

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Other answers:

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

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