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=$2700; Emily=$????

- anonymous

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- 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

@hartnn @ganeshie8 @primeralph

- anonymous

@Callisto

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## More answers

- anonymous

@mayankdevnani

- anonymous

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

- anonymous

what?

- anonymous

nope, they help me a lot

- anonymous

@satellite73

- 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

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

- hartnn

you have those sum formulas ?

- hartnn

and did you get that long explanation ? :P

- anonymous

why a1=1?

- 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

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

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

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

oh, it seems easy. Let me try

- anonymous

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

- 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

uhh.

- anonymous

0.000256???!

- hartnn

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

- 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

430?

- hartnn

yes, i get 430 too :)

- anonymous

thanks. and part b Emily?

- hartnn

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

- hartnn

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

- anonymous

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

- anonymous

thanks*

- hartnn

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

- anonymous

ah ha, i got it.

- 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

not total amount, IN THAT YEAR

- anonymous

okay..

- hartnn

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

- anonymous

n'th term

- hartnn

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

- anonymous

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

- 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

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

- hartnn

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

- anonymous

yes

- hartnn

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

- anonymous

3215...

- hartnn

is that the answer in your book ?
difference = 7700-3215

- anonymous

the answer is 635

- hartnn

i am getting 3850 for bobby

- hartnn

3850-3215=635

- hartnn

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

- anonymous

why 2 times 100?

- hartnn

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

- anonymous

uh? why? it's fixed

- anonymous

?

- hartnn

yeah, thats the general formula for sum...

- anonymous

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

- hartnn

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

- hartnn

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

- anonymous

ah ha, thanks

- anonymous

3215...
thank you so much

- hartnn

welcome ^_^

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