Question

Monthly = P (1 + r/12)12 = (monthly compounding)
Karan starts saving for a down-payment on a house by making regular payments at the
end of each month into an account that earns 8.4% / a interest, compounded monthly.
Determine the amount of each monthly payment needed to accumulate $10 000 in three
years.

@terenzreignz

Answer 1

Let's see...

Answer 2

Okay, so r = 8.4% = 0.084 right?

Answer 3

yes

Answer 4

But then again, regular monthly payments are made... looks like this is an annuity.
Am I right, @UnkleRhaukus ?
(sorry, money isn't really my thing :D )

Answer 5

yeah its annuity, its not my thing either aha

Answer 6

Okay, so we have a formula for an annuity
\[\huge a_{{n}}=\frac{1-(1+i)^{-n}}{i}\] right?

Answer 7

Where i is the effective rate of interest per period.

Answer 8

Catch me so far?

Answer 9

Sorry, \(\large a_n\) is the present value of an annuity that pays 1 per period.

Answer 10

yeah I'm following

Answer 11

But then again, we are given the accumulated value...
\[\huge s_n = \frac{(1+i)^n-1}{i}\]

Answer 12

So... we need to find how much she puts in every month, right?

Answer 13

yes that is correct

Answer 14

Okay, the accumulated value at the end of n-periods of a deposit of x per period is given by
\[\huge x \cdot s_n\]

Answer 15

Now, it is given that
\[\huge x\cdot s_n = 10000\]

Answer 16

Let's take this per month, shall we?

Answer 17

After all, she is depositing on a monthly basis.

Answer 18

okay

Answer 19

so... if it's 3 years, how many months is that?

Answer 20

sorry, 36

Answer 21

n=36 right?

Answer 22

yeah, 36.
Now what's the effective rate of interest per month?

Answer 23

0.084

Answer 24

No... that's for 12 months, remember? :P

Answer 25

hmm

Answer 26

the interest per month is 1/12 of the annual compounded interest :)

Answer 27

do i use r=e-1 to find it

Answer 28

i'm lost

Answer 29

No, the interest per month is simply
\[\huge \frac{0.084}{12}=0.007\]

Answer 30

thats what i did originally but i confused myself ...

Answer 31

well, don't...
\[\huge x \cdot s_n = 10000\]
\[\huge s_n = \frac{(1+i)^n-1}{i}\]
\[\huge i=0.007\]\[\huge n =36 \ (months)\]

Get to it, champ :)

Answer 32

aha thank you :)

Answer 33

10 000=(1+0.007)36-1
10 000= (1.007)36-1
10 000= 35.25 -1 ? am i doing this right so far?

Answer 34

you forgot to divide by i
And where's the x?

Answer 35

oops

Answer 36

k i confused myself

Answer 37

10 000 = 35.25/0.007 ?

Answer 38

Be careful with your applications of the operations...
What is
\[\huge 1.007^{36}\]?

Answer 39

1.007 to the power of 36

Answer 40

yeah, evaluate it... using your calculator.

Answer 41

1.28547

Answer 42

And use that.

Answer 43

\[\huge s_n = \frac{(1+i)^n-1}{i}\]

Answer 44

I have to go now, ok?
Good luck, sorry I couldn't stay longer...
-----------------------------------------------
Terence out

Answer 45

thank you so much for your help !