Question

Help with advanced algebra with financial applications. I want to see if I am right.

Answer 1

Salina purchased a dining room set for $1,730 using a 12-month deferred payment plan with an interest rate of 23.45%. She did not make any payments during the deferment period. What will the total cost of the dining room set be if she must pay off the dining room set within two years after the deferment period?

I got $2,182.26

Answer 2

are they monthly payments?

Answer 3

Yes, they are monthly payments.

Answer 4

you have period in which no payments are made, but the interest is adding on.

A = P(1+r/n)^(nt)

A = 1730(1+.2345/12)^(12*1) is the initial cost to start paying off right?

Answer 5

Well for this I used a different formula.

Answer 6

what did you use?

Answer 7

2,182.26 is the amount that she is going to finance, the total cost is the sum of her payments

Answer 8

\[M=B*\frac{ i(1+i)^nt }{ (1+i)^nt -1 }\]

Answer 9

her starting balance is 1730, this accrues to 2182.26 during the defered period

then she has 2 years to pay off the balance.

Answer 10

i use my own formula, seems to make more sense to me is all:
\[B_n=B_ok^n-P\frac{1-k^n}{1-k}\]
when B_n=0 the balance is paid off
\[0=B_ok^n-P\frac{1-k^n}{1-k}\]
\[B_ok^n=P\frac{1-k^n}{1-k}\]
\[B_ok^n\frac{1-k}{1-k^n}=P\]
and we want 24 payments which will be 2 years, n=24 k is the compounding stuff (1+.2345/12)
\[24*B_ok^{24}\frac{1-k}{1-k^{24}}=24P,k=(1+.2345/12)\]

Answer 11

this gets us a total cost of 24 payments which amounts to: 2754.73

http://www.wolframalpha.com/input/?i=24*2182.26k%5E%2824%29%281-k%29%2F%281-k%5E%2824%29%29%2Ck%3D%281%2B.2345%2F12%29

Answer 12

Ok. Could you explain to me how you got it?

Answer 13

we agree that the total cost will be the sum of the payments right?

Answer 14

yes

Answer 15

I got 2,754.73 after I used my equation again only this time I changed the original 1730 to 2,183.83.

Answer 16

lets start with a balance, and start making payments:

M0 = B0

M1 = B0 k - P

M2 = (B0 k - P)k - P
= B0 k^2 - Pk - P

M3 = (B0 k^2 - Pk - P)k - P
= B0 k^3 - Pk^2 - Pk - P


Mn = B0 k^n - Pk^(n-1) - P(n-1) - ... - P

now, what we have here is an original balance that is compounding, and a set of payments that follow a geometric progression so this simplifies to:

Mn = B0 k^n - P(1-k^n)/(1-k)

the rest is just solving for the length of time it takes to get our balance to 0

Answer 17

I have never used this type of equation. What do the letters mean so I can better understand this. Please.

Answer 18

we can work this arrangement into the formula you use, but i find it simpler to use it at this stage

Answer 19

B0, the original balance

P, payment amount

k is a compounding interest (1+r/s), s being how many times a year we are divvying up the interest

and n is the number of periods/payments that it takes

i used M for month but it really tells us the balance for that month

Answer 20

Ok.

Answer 21

i could never work the formula in the texts, so i simply made my own :)

Answer 22

either way, 2,754.73 or thereabouts is what we should have

Answer 23

Ok. I get it now

Answer 24

I see. I got the equation as I did it by myself. Thanks for clearing it up!