Hi there, I'm having a bit of trouble with this question and any help would be appreciated!

A loan of $5600 is to be repaid at 9% compounded annually by making 10 payments at the end of each quarter. Each of the last six payments are two times the amount of each of the first four payments. What is the size of each payment? 2. Originally Posted by reardear Hi there, I'm having a bit of trouble with this question and any help would be appreciated! A loan of$5600 is to be repaid at 9% compounded annually by making 10 payments at the end of each quarter. Each of the last six payments are two times the amount of each of the first four payments. What is the size of each payment?
WHAT trouble are you having?

You need to do these steps ( let p = payment at start) :

1: convert 9% annual to its equivalent quarterly rate:
(1 + i)^4 = 1.09 ; solve for i which will be the quarterly rate.

2: get the future value of 5600: f = 5600(1 + i)^10

3: get future value of first 4 payments:
u = p[(1 + i)^6][(1 + i)^4 - 1] / i

4: get future value of last 6 payments: v = 2p[(1 + i)^6 - 1] / i

So we have: f = u + v : solve for p

3. Originally Posted by reardear
Hi there, I'm having a bit of trouble with this question and any help would be appreciated!

A loan of \$5600 is to be repaid at 9% compounded annually by making 10 payments at the end of each quarter. Each of the last six payments are two times the amount of each of the first four payments. What is the size of each payment?
You should abandon your trouble and practice BASIC PRINCIPLES. You will never struggle with the "right formula" again.

i = 0.09

j = (1+i)^(1/4) - 1

v = 1/(1+j)

That's all you need. Now build it.

5600 = Pmt(v + v^2 + v^3 + v^4 + 2(v^5 + v^6 + v^7 + v^8 + v^9 + v^10))

$5600 = Pmt\cdot\left[\left(\frac{v-v^{5}}{1-v}\right) + 2\cdot\left(\frac{v^{5} - v^{11}}{1-v}\right)\right] = \frac{Pmt\cdot v}{1-v}\cdot\left[\left(1-v^{4}\right)+2\cdot\left(v^{4}-v^{10}\right)\right]$

Simplify a little more and GO!

4. Originally Posted by Denis
WHAT trouble are you having?

You need to do these steps ( let p = payment at start) :

1: convert 9% annual to its equivalent quarterly rate:
(1 + i)^4 = 1.09 ; solve for i which will be the quarterly rate.

2: get the future value of 5600: f = 5600(1 + i)^10

3: get future value of first 4 payments:
u = p[(1 + i)^6][(1 + i)^4 - 1] / i

4: get future value of last 6 payments: v = 2p[(1 + i)^6 - 1] / i

So we have: f = u + v : solve for p
1) Okay, so I get i = 0.021778181
2) 6946.31
3) [5600(1.137993409)(.009)] / i = 26335.93
4) [11500(.137993409)] / i = 70966.73

Have I done it correctly so far? And since there's no specified dates for these payments, would I simply divide #3 by 4 and #4 by 6?

Edit: Oh, and I'm not sure where p comes from when you ask to solve for it at the end

5. Originally Posted by reardear
1) Okay, so I get i = 0.021778181
2) 6946.31

Edit: Oh, and I'm not sure where p comes from when you ask to solve for it at the end
Both correct...but not the rest!
I told you the p represented the payment: here's my post again:
.................................................. ........................................
You need to do these steps ( let p = payment at start) :

1: convert 9% annual to its equivalent quarterly rate:
(1 + i)^4 = 1.09 ; solve for i which will be the quarterly rate.

2: get the future value of 5600: f = 5600(1 + i)^10

3: get future value of first 4 payments:
u = p[(1 + i)^6][(1 + i)^4 - 1] / i

4: get future value of last 6 payments: v = 2p[(1 + i)^6 - 1] / i

So we have: f = u + v : solve for p
.................................................. ........................................

I've given you the "breakdown" to use to do the payment calculation;
I'm using the "future value" approach:
TKHunny showed you the "present value" approach:
did you not follow his approach also?

OK, from what I gave you above:
i = 1.09^(1/4) - 1
Let k = 1 + i

p(k^4 -1)(k^6) / i + 2p(k^6 - 1) / i = 5600(k^10)

p(k^4 -1)(k^6) + 2p(k^6 - 1) = 5600(i)(k^10)

p[(k^4 -1)(k^6) + 2(k^6 - 1)] = 5600(i)(k^10)

p = 5600(i)(k^10) / [(k^4 -1)(k^6) + 2(k^6 - 1)]

Now substitute k = 1 + i in above and you'll get p = 399.7768...

If you solve TKHunny's for "Pmt", you'll get same result.

NOTE: to keep TK in good humor, use 399.77, not 399.78

6. Originally Posted by tkhunny
5600 = Pmt(v + v^2 + v^3 + v^4 + 2(v^5 + v^6 + v^7 + v^8 + v^9 + v^10))
This might be more enlightening

5600 = Pmt*(v + v^2 + v^3 + v^4 + v^5 + v^6 + v^7 + v^8 + v^9 + v^10) + Pmt*(v^5 + v^6 + v^7 + v^8 + v^9 + v^10)

5600 = Pmt*v*(1 + v + v^2 + v^3 + v^4 + v^5 + v^6 + v^7 + v^8 + v^9) + Pmt*(v^5)*(1 + v + v^2 + v^3 + v^4 + v^5)

This second one is very instructive. Look how it rewords the description. Rather than increasing the payment, it suggests a level payment stream for the entire period, increased by a different payment stream deferred five periods!

5600 = Pmt*v*[(1 + v + v^2 + v^3 + v^4 + v^5 + v^6 + v^7 + v^8 + v^9) + (v^4)*(1 + v + v^2 + v^3 + v^4 + v^5)]

$5600 = Pmt\cdot v\cdot\left[\left(\frac{1-v^{10}}{1-v}\right) + v^{4}\cdot \left(\frac{1-v^{6}}{1-v}\right)\right]$

$5600 = Pmt\cdot \frac{v}{1-v}\cdot\left[\left(1-v^{10}\right) + v^{4}\cdot \left(1-v^{6}\right)\right]$

Really, a little algebra and the world is a better place. Unique answers don't care how you find them.

7. Thank you both for the help! At first I couldn't recognize the formulas you were using, but I realized the ones I'm normally used to seeing are just slightly different and have different representations. Thanks again!

8. I hope you noticed that I didn't use a FORMULA at all! I built the solution myself!!

This is not pat-me-on-the-back day. It's a "Basic Principles" approach that will save you in the stickiest of situations.