Sequences and series

**Coach** · March 25th 2008, 10:24 AM

Dear forum members, could you please verify that my calculations are correct

A person takes a two million euro loan from a bank with an annual fixed interest of 6%. He plans to pay it back in 10 years. In order to be able to complete his goal, how much must he pay per month?

q=1,005 (the coefficient that will tell us the interest added to the loan)

n=10 (the years we want to spend on paying the loan)

a= 2 000 000 (the amount of money taken)

b= ? (the amount of money we have to pay per month)

$q^n*a -\frac{b(1-q)}{1-q}\leq0$

Oh yeah, and if anyone was wondering I used geometric sum for the fraction part.

I got for my answer for solving b, about 20 5541.1457€ , but I highly doubt this is correct.

So if someone could check it, I would be very grateful.

**Soroban** · March 25th 2008, 12:12 PM

Hello, Coach!

Sorry, your answer is incorrect . . .

A person takes a $2,000,000 loan from a bank with an annual fixed interest of 6%.
He plans to pay it back in 10 years.
In order to be able to complete his goal, how much must he pay per month?

There is an Amortization Formula: . $A \;=\;P\frac{i(1+i)^n}{(1+i)^n-1}$
. . where: . $\begin{Bmatrix}A & = &\text{periodic payment} \\ P &=& \text{principal} \\ i &=& \text{periodic interest rate} \\ n&=&\text{number of periods} \end{Bmatrix}$

We have: . $P \:=\:2,000,000,\;i \:=\:\frac{6\%}{12} \:=\:0.005,\;n \:=\:120$

Hence: . $A \;=\;2,000,000\cdot\frac{0.005\cdot1.005^{120}}{(1 .005)^{120}-1} \;=\;22,204.10039$

Therefore, his monthly payment will be: . $\$22,204.10$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The derivation of the Formula is lengthy but quite elementary.

I think I can recreate the steps . . . if anyone is interested.

**Coach** · March 25th 2008, 12:14 PM

Originally Posted by Soroban

Hello, Coach!

Sorry, your answer is incorrect . . .

There is an Amortization Formula: . $A \;=\;P\frac{i(1+i)^n}{(1+i)^n-1}$
. . where: . $\begin{Bmatrix}A & = &\text{periodic payment} \\ P &=& \text{principal} \\ i &=& \text{periodic interest rate} \\ n&=&\text{number of periods} \end{Bmatrix}$

We have: . $P \:=\:2,000,000,\;i \:=\:\frac{6\%}{12} \:=\:0.005,\;n \:=\:120$

Hence: . $A \;=\;2,000,000\cdot\frac{0.005\cdot1.005^{120}}{(1 .005)^{120}-1} \;=\;22,204.10039$

Therefore, his monthly payment will be: . $\$22,204.10$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The derivation of the Formula is lengthy but quite elementary.

I think I can recreate the steps . . . if anyone is interested.

Oh, thank you so much!!!!!!!!!!!!!

Well, I almost got it

**Coach** · March 25th 2008, 12:17 PM

However, if you could please explain why my approach did not work?
Is it wrong to make it an inequality, and use the formulas for solving a geometric sum?

**Coach** · March 25th 2008, 12:52 PM

And when I try checking it

I get that the final amount with interets added that he would have to pay would be $1,05^{10}*2000 000$ and that does not equal the monthly payment times 120?

Why is that? Shouldn't they equal each other.

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Sequences and series, a financial application with interest rates, please help me

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