# Thread: Stuck with geometric series question: £50,000 is borrowed for a house....

1. ## Stuck with geometric series question: £50,000 is borrowed for a house....

Can't fathom out geometric series question despite knowing the formulas!

The question: £50,000 is borrowed for a house. At end of each year 8% interest is charged on amount still owed. A fixed amount (£P) is payed at the end of each year so that the amount owed after first year is: (50,000 x 1.08) - P

so after third payment it would be (50,000 x 1.08^3) -P(1+1.08+1.08^2). (n less than or equal to 30)

Formula for nth payment would then be: (50,000 X1.08^n) - P(1+1.08+1.08^2.......1.08^n-1). I understand this SO FAR.

The question then asks: ' Use the formula for the sum of a geometric progression to simplify the formula and use this to find the amount £P with n=30 giving answer to nearest penny.

As these are finite divergent series I thin I should use a(r^n-1)/(r-1) but not sure how unless I use this:

[50,000(1.08^30-1)/0.08] - [P(1.08^29-1/0.08)] ??? Not sure what to do here!

Any help-much thanks!

2. Originally Posted by Simonsky
The question: £50,000 is borrowed for a house. At end of each year 8% interest is charged on amount still owed. A fixed amount (£P) is payed at the end of each year so that the amount owed after first year is: (50,000 x 1.08) - P

so after third payment it would be (50,000 x 1.08^3) -P(1+1.08+1.08^2). (n less than or equal to 30)

Formula for nth payment would then be: (50,000 X1.08^n) - P(1+1.08+1.08^2.......1.08^n-1). I understand this SO FAR.

The question then asks: ' Use the formula for the sum of a geometric progression to simplify the formula and use this to find the amount £P with n=30 giving answer to nearest penny.

As these are finite divergent series I thin I should use a(r^n-1)/(r-1) but not sure how unless I use this:

[50,000(1.08^30-1)/0.08] - [P(1.08^29-1/0.08)] ??? Not sure what to do here!
You're missing a pair of parentheses in that last expression, but otherwise it looks good: [50,000(1.08^30-1)/0.08] - [P(1.08^29-1)/0.08]

They seem not to have stated one important thing: I think they meant to say that the loan is paid off in 30 years, so that your expression should equal 0 (nothing still owed at that time). That gives you an equation to solve. Does this sound like what they intended? Did you omit such a statement?

3. You're sure looking at it correctly!
FV of 50000 after n years = FV of annual annuity of p after n years

a = 50000
r = .08
n = 30
p = ?

a(1 + r)^n = p{[(1 + r)^n - 1] / r}

p = ? : your job

You should end up with p = 4431.3716...

Edit: darn it, sorry Dr.P, did not see your post; I'm assuming 30 annual payments...

4. Originally Posted by Dr.Peterson
You're missing a pair of parentheses in that last expression, but otherwise it looks good: [50,000(1.08^30-1)/0.08] - [P(1.08^29-1)/0.08]

They seem not to have stated one important thing: I think they meant to say that the loan is paid off in 30 years, so that your expression should equal 0 (nothing still owed at that time). That gives you an equation to solve. Does this sound like what they intended? Did you omit such a statement?

yes! The question was worded in a lot more detail so I thinned it out -thanks, you've helped me there!

Simon

5. Originally Posted by Simonsky
yes! The question was worded in a lot more detail so I thinned it out -thanks, you've helped me there!

Simon

The answer ( I caved in in the end) shows the modified formula as: 50,000 x 1.08^n - 25P/2(1.08^n-1)

Not sure how the 25/2 other than 12.5 x 0.08 =1

Any help appreciated.

6. Did you try finishing what either of us suggested?

Originally Posted by Dr.Peterson
You're missing a pair of parentheses in that last expression, but otherwise it looks good: [50,000(1.08^30-1)/0.08] - [P(1.08^29-1)/0.08]

They seem not to have stated one important thing: I think they meant to say that the loan is paid off in 30 years, so that your expression should equal 0 (nothing still owed at that time). That gives you an equation to solve. Does this sound like what they intended? Did you omit such a statement?
The implication of this is that you should solve (based on your formula)

[50,000(1.08^30-1)/0.08] - [P(1.08^29-1)/0.08] = 0

This amounts to solving

[50,000(1.08^30-1)/0.08] = [P(1.08^29-1)/0.08]

Now, I hadn't carefully checked your details, just that you were doing generally the right thing; I didn't note that where you previously had (50,000 X1.08^n) on the left, you changed it to [50,000(1.08^30-1)/0.08], as if you were summing those terms. And on the right, because the last term in the summation has exponent n-1 = 29, the formula should give 30, not 29. So the equation I should have told you to write was

50,000 x 1.08^30 = P(1.08^30-1)/0.08

Originally Posted by Denis
a(1 + r)^n = p{[(1 + r)^n - 1] / r}

You should end up with p = 4431.3716...
Plugging in the numbers, this says you are to solve

50000(1.08)^30 = p{[(1.08)^30 - 1] / .08}

This is the same as my corrected version above, based on your work.

Originally Posted by Simonsky
The answer ( I caved in in the end) shows the modified formula as: 50,000 x 1.08^n - 25P/2(1.08^n-1)

Not sure how the 25/2 other than 12.5 x 0.08 =1
This, with n=30, is identical to Denis's, since as you point out 25/2 = 1/0.08.

You had that division by 0.08 yourself; the only problem is that your version had a subtraction and a division by 0.08 on the left, too.

So, solve this equation, and see if you get what Denis got (or whatever your book says)!

7. In case you're interested in the standard "loan payment formula":
http://financeformulas.net/Loan_Payment_Formula.html

8. This, with n=30, is identical to Denis's, since as you point out 25/2 = 1/0.08.

You had that division by 0.08 yourself; the only problem is that your version had a subtraction and a division by 0.08 on the left, too.

So, solve this equation, and see if you get what Denis got (or whatever your book says)!

Thanks Dr. P that 25P/2 is now clear to me.......maths requires the ability to see the same thing from different angles and I'm still poor at that. Didn't someone say 'maths is the art of saying the same thing different ways'?

Help much appreciated..would not have done it on my own.

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