#### DenisB

Code:
YEAR   TRANSACTION   INTEREST      BALANCE
1     3000.00           .00      3000.00
2     3000.00        360.00      6360.00
3     3000.00        763.20     10123.20
3        ?              .00        ?
4     3000.00           ?          ?
5     3000.00           ?        8000.00
3000 is deposited annually for 5 years, interest rate = 12% annual.

At end of 5th year, after 5th deposit, there is exactly 8000 in the account.

For this to be possible, a transaction was made at end of the 3rd year.

What is the amount of that transaction?

d = 3000 (periodic deposit)
f = 8000 (ending amount)
i = .12 (periodic rate)
n = 5 (number of periods)
m = 2 (periods left after transaction
t = ? (transaction amount)

What is t in terms of d,f,m,n,i ?

Have fun!!

#### chiro

MHF Helper
Hey DenisB.

Hint: If you look at one term then the next term will have New Amount = Old Amount + Interest*[Old Amount] = [Old Amount]*(1+Interest). Note that this is for one term only.

You can specify this for multiple terms taking into account that a transaction will add some figure to the amount at the start of that term.

#### DenisB

Thanks Chiro...I made that up as a mini-challenge...
so I have the solution, of course...
Should I have posted elsewhere?

#### DexterOnline2

$$\displaystyle t=-8,815.80$$

#### DenisB

NO: -8815.802040816326530021...

You are hereby demoted to the mail room...

#### JeffM

Code:
YEAR   TRANSACTION   INTEREST      BALANCE
1     3000.00           .00      3000.00
2     3000.00        360.00      6360.00
3     3000.00        763.20     10123.20
3        ?              .00        ?
4     3000.00           ?          ?
5     3000.00           ?        8000.00
3000 is deposited annually for 5 years, interest rate = 12% annual.

At end of 5th year, after 5th deposit, there is exactly 8000 in the account.

For this to be possible, a transaction was made at end of the 3rd year.

What is the amount of that transaction?

d = 3000 (periodic deposit)
f = 8000 (ending amount)
i = .12 (periodic rate)
n = 5 (number of periods)
m = 2 (periods left after transaction
t = ? (transaction amount)

What is t in terms of d,f,m,n,i ?

Have fun!!
The problem should say that the 3000 is deposited at the end of the year (although you can figure that out from the attached table).

$f = d\left(\dfrac{(1 + i)^n - 1}{i}\right) + t(1 + i)^m \implies t(1 + i)^m = f - d\left(\dfrac{(1 + i)^n - 1}{i}\right) \implies$

$t = \left\{f - d\left(\dfrac{(1 + i)^n - 1}{i}\right)\right\} \div (1 + i)^m.$

Substituting in the numbers

$t = \left\{8000 - 3000\left(\dfrac{(1.12)^5 - 1}{0.12}\right)\right\} \div (1.12)^2 \approx -\ 8815.80.$

The negative indicates a withdrawal of course.

Now, denis, you really should check your answer before you turn anything in, but I'll let you do that as an exercise. And next time, please do as chiro asks, and don't expect us to do your homework. (Alternatively you could post in the math challenge forum. The challenge here is mostly to come up with the cleanest formula. My first effort was pretty kludgy.)

EDIT: And now I notice that dexteronline COMPLETELY missed the problem because he gave a numeric rather than algebraic answer.

Last edited:

#### DenisB

Forgiveth us, oh Father Jeffery

#### DexterOnline2

EDIT: And now I notice that dexteronline COMPLETELY missed the problem because he gave a numeric rather than algebraic answer.
@Sir Jeff

The odds of you finding the equation I used in finding $$\displaystyle t=8,815.80$$ are $$\displaystyle \frac{1}{\infty}=0$$ as there are an infinite number of equations that satisfy the relationship amongst the many variables that honorable Sir DenisB listed in his OP.

Your's or mine equation made use of future value at $$\displaystyle n$$

$$\displaystyle d\left(\dfrac{(1 + i)^n - 1}{i}\right) + t(1 + i)^{m} - f = 0$$

We could have used the following equation for present value at $$\displaystyle 0$$

$$\displaystyle d\left(\dfrac{1 - (1 + i)^{-n}}{i}\right) + t(1 + i)^{m-n} - f(1 + i)^{-n} = 0$$

We could have also used the following equation for intermediate value at any time horizon between $$\displaystyle k=0$$ and $$\displaystyle k=n$$ therefore infinite such time slots

$$\displaystyle d (1 + i)^{k} \left(\dfrac{1 - (1 + i)^{-n}}{i}\right) + t(1 + i)^{k}(1 + i)^{m-n} - f(1 + i)^{k}(1 + i)^{-n} = 0$$

We could have also used the following equation for horizon value using any non-negative interest rate $$\displaystyle h_r$$ at any time horizon between $$\displaystyle h=0$$ and $$\displaystyle h=n$$ therefore infinite such time slots

$$\displaystyle d (1 + h_r)^{h} \left(\dfrac{1 - (1 + i)^{-n}}{i}\right) + t(1 + h_r)^{h}(1 + i)^{m-n} - f(1 + h_r)^{h}(1 + i)^{-n} = 0$$

All those would results in a value of $$\displaystyle t=8,815.80$$

For the last two equations, we may have also gone beyond the horizon $$\displaystyle k=0$$ and $$\displaystyle k=n$$ or $$\displaystyle h=0$$ and $$\displaystyle h=n$$ such as $$\displaystyle -\infty<k<\infty$$ or $$\displaystyle -\infty<h<\infty$$ and still get the same value of $$\displaystyle t=8,815.80$$. Those would then be beyond horizon values in contrast to present, intermediate, horizon and future values.

Somewhere some other place, Dexter refers to infinite financial formulas on a single grain of sand on Long Beach in California. Here was a snapshot of such one baby infinity compare that to Grand infinity, the sum of all baby infinities.

1 person