Question

an individual saves $5000 a bank account at the beginning of each year for 10 years. No further saving or withdrawals are made from the account. determines the total amount saved if the annual interest rate is 8% compounded.

(a) annually

(b) semi-annually

Answer 1

i think its annualy

Answer 2

i should calculate both of them

Answer 3

5000(1+8/100)^10 ?

Answer 4

Really? This s NOT different than those e did before IF you understad teh basic principles.

i = 0.08

RA = 1 + i = 1.08 -- Annual Accumulation Factor
RS = 1 + i/2 = 1.04 -- Semi-Annual Accumulation Factor

Annual Accumulation

\(5000(RA^{10} + RA^{9} + ... + RA)\) -- Done.

Semi-Annual Accumulation

\(5000(RS^{20} + RS^{18} + ... + RS^{2})\) -- Done.

Seriously, forget the formulas and learn to draw the map!

Answer 5

what grade is this?

Answer 6

annual it is 5000(1.08^10-1/1.08-1) =7243

Answer 7

first year at university

Answer 8

oops then i shouldnt be helping :)

Answer 9

annual sorry it is 72432
and semi annual 5000(1.04^20-1/1.04-1)=148890

Answer 10

No good. You are thinking about formulas and not about basic principles. If you think about formulas, then every chapter or seciont will seem new and will be increasingly confusing. If you think about basic principles, you will never need t obe confused again.

\(5000(RA^{10} + RA^{9} + ... + RA) = 5000\frac{RA - RA^{11}}{1-RA}\)

There are other wayso write this formula, b none more useful if yojust created it yourself and you have ultimate confiden that it correct.

Answer 11

ohh kk 10 + first term kk understand!

Answer 12

You have to start thinking about it more. You seem to be substantially guessing.

For starters, accumulating 5,000 annually for 10 years is AT LEAST 50,000. Do not EVER accept a values less than 50,000. Your "7243" cannot possibly be even close. It should make sense before you think it is the answer.

Answer 13

annual =7190
but you can calculate the semi annual day ? if it start from (r^2 >r^20

Answer 14

how it can 5000 time 10 if you have interest 8%?

Answer 15

72432 is no good. Are you rounding badly again?

7232 is no good. You are making payments at the END of the year.

It is NOT 5000*10 = 50000. This is just a reality check. If you get 7243, that clearly is wrong. The answer MUST be greater than 50,000.

Answer 16

You're still just guessing.

One step at a time.

i = 0.08 -- Annual interest rate
r = 1+i = 1.08 -- Annual accumulation factor

"5000 at the beginning of each of 10 years"

#1 - 5000*r^10
#2 - 5000*r^9
#3 - 5000*r^8
...
#10 - 5000*r^1

Add them up: 5000*r^10 + 5000*r^9 + 5000*r^8 + ... + 5000*r^1

A little algebra: 5000*(r^10 + r^9 + r^8 + ... + r^1)

A little algebra: \(5000\dfrac{r - r^{11}}{1-r}\)

A little calculator work

r^11 = 2.3316389971
r - r^11 = -1.2516389971
1 - r = -.08

Final answer: 5000((-1.2516389971)/(-0.08)) = 5000*15.6454874638 = 78227.44

There is no need to guess.

Answer 17

Calculator:
Type = Beginning of Period

N = 10

i = 0.08 or 8.00 -- However your calculator expects to see this value.

Pmt = 5000 or maybe -5000 -- Again, make sure you understand what your calculator wants.

PV = 0

Calculate FV

Answer 18

i just confusing because i did R^11-1/0.8*5000 i did for annual because i thought that first term should 1.08^0 =1> 1.08^10

Answer 19

We need to back up a little bit. Have you studied the "Order of Operations"?

Answer 20

no

Answer 21

It is usually very early on - possibly before algebra. Are you sure you haven't seen it? It usually comes out like "PEMDAS".

Answer 22

no usually interest coumpound and geomectric serie only

Answer 23

That's okay. It is not as prominent in all cultures. U.S. math texts usually harp on it. It is a way to write more clearly in a standardized fashion so that the things you write can be more frequently and more clearly understood. For example, when you write "R^11-1/0.8*5000 ", how do I know the order in which you did this calculation? Usually, some judicious parentheses will clear it up.

(R^11-1)/0.8*5000

Sometimes spacing makes it more clear.

(R^11 - 1)/0.8*5000

Sometimes order makes it more clear

5000*(R^11 - 1)/0.8

See what I mean?

Answer 24

yes know i understand we in Lebanon that you should see how we studied now i understand that

Answer 25

1.33163/0.8*5000=83227

Answer 26

Good. This is just encouragement to try to write more clearly. Make sure it says what you want it to say. Remember that we are not inside your head. -)

Okay, enough of that. Back to the problem...

Your statement "5000*(R^11 - 1)/0.8" has a couple of problems.

1) You have 0.8 in the denominator. No good. Maybe 0.08?
2) You cannot have R^11 with 1 in the numerator and the interest rate in the denominator unless you have 11 payments. You have only 10.
3) All this leads me to believe you are still kind of just guessing. You're not quite sure which formula to use.

How are we doing?

Answer 27

"1.33163/0.8*5000=83227" This s a good example of what I was talking about. You WROTE "0.8" but you MEANT and USED "0.08". Write what you mean!

Answer 28

look what i did
r+R^1.....r^10 so 11 term
i did by my head calculation 8/100=0.8 i see it that it is wrong it is 0.08
so that why i did 5000(r^11-1=r^0)/0.08 that i a did

Answer 29

No. r + R^1 <== What does that mean? You have an extra term.

Answer 30

ahh kk so it r^10-1?/0.08

Answer 31

if it 10 year r^10-1/0.08*5000

Answer 32

Closer, but still no. That would be correct for payments at the END of the period. You have payments at the BEGINNING of each period.

Answer 33

hard to understand term begenning and end period ^^

Answer 34

BEGINNING just has one more accumulation period than END.

End: \(\dfrac{r^{N} - 1}{i}\)

Beginning: \(r\cdot\dfrac{r^{N} - 1}{i} = \dfrac{r^{N+1} - r}{i} = \dfrac{r^{N} - 1}{i\cdot v} = \dfrac{r^{N} - 1}{d}\)

You may see any or all of these other forms.

Answer 35

d? what is it

Answer 36

i*v ?

Answer 37

It is a standard definition.

\(v =\dfrac{1}{1+i}\)

Answer 38

we didnt take standard definition

Answer 39

Another standard definition

\(d = i\cdot v = \dfrac{i}{1+i}\)

You should be familiar with these. You WILL encounter them.

Answer 40

but we didnt studied yet i can use standar for my homework...

Answer 41

You can use what you can define.

Answer 42

The semi-annual problem is almost the same. You just have to redefine r and i.

For annual, we had r = 1+i = 1.08

For semi-annual, still with annual payments, we have \(r = \left(1 + \dfrac{0.08}{2}\right)^2\).

This makes i = 0.0816 for the semianual calulation.

Answer 43

kk that make send sens i try to understand i how find annual anwser without using standar formual we didnt learned

Answer 44

annual
i found 5000(1.08)(1.08^10-1/1.08-1)

Answer 45

I believe that, except for the lack of clarity.
I would have preferred 5000(1.08)(1.08^10 - 1)/(1.08-1)

You should be able to solve the semi-annual one using i = 0.0816

Answer 46

5000(1.04)(1.04^20-1)/(1.04-1)

Answer 47

No. That would be good for 10 years of semi-annual payments. That is NOT what you have. You have 10 years of ANNUAL payments that are compounded semi-annually.

Answer 48

semie annual so 5000(1+8/2/100)^10*2

Answer 49

when we use geometrci serie it should like i did !

Answer 50

You need the equivalent annual interest related to your semi-annual compounding. Once you have that, it is EXACTLY like the first problem in this pair.

Answer 51

One ALWAYS uses the Geometric Series, otherwise the theory doesn't exist.. Do not separate what you are seeing from the Geometric Series fundamental development.

Answer 52

mmmm...

Answer 53

Your compounding period is different from your payment period. It takes a little more thought. Remember this?

\(i = 0.08\)

\(RS = 1 + i/2 = 1.04 \)

\(5000(RS^{20} +RS^{18} +...+ RS^{2})\) -- Done.

\(5000\cdot\dfrac{RS^{20} - 1}{RS^{2} - 1}\cdot RS^{2}\) -- Also done.

Answer 54

it should be rs^2? because it semi anual
but why not 5000(1.04)?

Answer 55

Payments are only every other compounding period. There is not a payment every ½ year.

Answer 56

ohhh kk thank you so much

Answer 57

Time 0 5000
Time ½ 5000*(1.04)
Time 1 (5000*1.04*1.04 + 5000)

Answer 58

understsand^^ finaly