Sequence and series (1 Viewer)

fullonoob · Jan 21, 2010

A $2000 loan is offered at 18% pa interest, charged monthly, over 4 years. If no repayments need to be paid for the first 3 months, find the amount of each repayment. How much will be paid back altogether?

Ifyou get 64.25 and 2891.25 post your working out here please.
Otherwise im pretty sure the question is wrong cos i got a diff answer to it. Tell answer if you get a diff answer
THX

~fullonoob

fullonoob · Jan 22, 2010

A loan of $6000 over 5 years at 15% p.a. interest, charged monthly, is paid back in 5 yearly instalments. How much is each instalment, and how much money is paid altogether?

ZOMG read this question. It contradicts itself or i dont understand. I think you do double GP or something. Answer is 1835.68; 9178.40.
LOL HELP

Lukybear · Jan 22, 2010

fullonoob said:
A $2000 loan is offered at 18% pa interest, charged monthly, over 4 years. If no repayments need to be paid for the first 3 months, find the amount of each repayment. How much will be paid back altogether?

$T_{1}=2000 \times (1+\frac{0.18}{12})^3 = 2091.35 \rightarrow A$

$n=4\times 12 - 3 = 45$
$T_{1}=2000 \times (1+\frac{0.18}{12})^3 = 2091.35 \rightarrow A$
$(1+\frac{0.18}{12}) \Rightarrow B$
$P_{1} = A\times B - R\\ P_{2} = A\times B^2 - R \times B - R\\P_{3} = A\times B^{3} - R\times B^{2} - R\times B - R\\...\\ P_{45} = A\times B^{45} - R\times B^{44} - ... - R\\ \break \text{at } P_{45} = 0\\ 0 = AB^{45} - RB^{44} - ... - R\\ 0 = AB^{45} - R(1+B^1+B^2+B^3...B^{44})\\ \text{R can now be solved using } R\frac{B^{45}-1}{B-1} = R(1+B^1+B^2+B^3...B^{44})\\ \break R=64.25\\ \text{Hence Total Payment} = 2891.25$

addikaye03 · Jan 22, 2010

Lukybear said:
$T_{1}=2000 \times (1+\frac{0.15}{12})^3 = 2091.35 \rightarrow A$

$n=4\times 12 - 3 = 45$
$T_{1}=2000 \times (1+\frac{0.15}{12})^3 = 2091.35 \rightarrow A$
$(1+\frac{0.18}{12}) \Rightarrow B$
$P_{1} = A\times B - R\\ P_{2} = A\times B^2 - R \times B - R\\...\\ P_{45} = A\times B^45 - R\times B^44 - ... - R\\ \break \text{Solving for R Gives}\\ \break R=64.25\\ \text{Hence Total Payment} = 2891.25$

Just a tip, prob no big deal but if a Q says find a statement for P_45 or whatever, you have to establish a pattern. So that means you need to show T1, T2 AND T3. Just showing T1 and T2 doesn't technically establish a pattern. ie. Find next number in sequence: 2, 4, _. Could view it multiple of 2 (so 6), or could be (T_n-1)^2 (so 16).

Just a small point, other than that, nice solution

fullonoob · Jan 22, 2010

Lukybear said:
$T_{1}=2000 \times (1+\frac{0.15}{12})^3 = 2091.35 \rightarrow A$

$n=4\times 12 - 3 = 45$
$T_{1}=2000 \times (1+\frac{0.15}{12})^3 = 2091.35 \rightarrow A$
$(1+\frac{0.18}{12}) \Rightarrow B$
$P_{1} = A\times B - R\\ P_{2} = A\times B^2 - R \times B - R\\...\\ P_{45} = A\times B^45 - R\times B^44 - ... - R\\ \break \text{Solving for R Gives}\\ \break R=64.25\\ \text{Hence Total Payment} = 2891.25$

hmm thanks but how do you get the 0.15 in the beginning. Btw could you use M instead of A and B, it makes me even more confused xD
Kinda dont get it still...could you show the full working out - without skipping steps please.

Lukybear · Jan 22, 2010

Um, sorry, its 0.18. You can use whatever. I just didnt wanna type it again and again so i juts used a sub.

Yea i get your point addikaye03. My fault. Im too lazy and careless, unlike you

.

Lukybear · Jan 22, 2010

fullonoob said:
A loan of $6000 over 5 years at 15% p.a. interest, charged monthly, is paid back in 5 yearly instalments. How much is each instalment, and how much money is paid altogether?

ZOMG read this question. It contradicts itself or i dont understand. I think you do double GP or something. Answer is 1835.68; 9178.40.

$n=5\times 12=60\\ \text{Rate Increase} = (\frac{0.15}{12}+1)^{12}\Rightarrow r \\ \text{Since percentage rate is charged monthly, with repayment only yearly. Hence compound interest of month will apply for a year}\\ \text{Rest ensues}\\ P_{1} = 6000r - R \\ P_{2} = 6000r^2 - Rr -R\\ P_{3} = 6000r^3 - Rr^2 - Rr - R\\ ...\\ P_{5} = 6000r^5 - Rr^4 - ... - r\\ \text{Solving for R using } S_{n} = \frac{a(r^n-1)}{r-1} \text{will give answer.}$

fullonoob · Jan 23, 2010

LOL THANKS A LOT LUKY

i did the exact same thing xDD
But i did it in 2 dp so when i got the answer it was like $20-30 off and i was like wtf

Then i did exact values and got the right answer :mad1:
me FAIL

fullonoob · Jan 23, 2010

1) Get rich Bank offers a mortgage at 13.5% pa. over 10 years and capital bank offers a mortgage at 10.5% pa. over 25 years.
Find the amount of repayment to each bank on a loan of $80000.

2) Joan's mother puts $300 into an account at the beginning of each year to pay for Jane's education in 5year's time. If interest of 6% pa. is paid quarterly, how much money will Jan'es mother have at the end of 5 years?

ROFLMAO sequence and series marathon here with me asking all the questions
I'll give you rep for the heck of it - thanks a lot

fullonoob · Jan 24, 2010

Question not answered still?

Help please

gwc · Nov 30, 2016

fullonoob said:
1) Get rich Bank offers a mortgage at 13.5% pa. over 10 years and capital bank offers a mortgage at 10.5% pa. over 25 years.
Find the amount of repayment to each bank on a loan of $80000.

Get Rich Bank
P = $80 000
r = 0.075/12 = 0.0065
n = 12x10 = 120

A₁ = 80 000(1.0065) - M
A₂ = 80 000(1.0065)² - M(1+1.0065)
A₃ = 80 000(1.0065)³ - M(1+1.0065+1.0065²)
A_n = 80 000(1.0065)ⁿ - M(1+1.0065+1.0065²+....+1.0065^n-1)
A₁₂₀ = 80 000(1.0065)¹²⁰ - M(1+1.0065+1.0065²+....+1.0065¹¹⁹)
0 = 80 000(1.0065)¹²⁰ - M(1+1.0065+1.0065²+....+1.0065¹¹⁹)
M = 80 000(1.0065)¹²⁰ / ((1.0065¹²⁰-1)/1.0065-1))
= $949.61

Capital Bank
P = $80 000
r = 0.055/12
n = 25x12 = 300

A₁ = 80 000(1+ 0.55/12) - M
A₂ = 80 000(12.055/12)² - M(1+12.055/12)
A₃ = 80 000(12.055/12)³ - M(1+12.055/12+(12.055/12)²)
...
A_n = 80 000(12.055/12)ⁿ - M(1+ 12.055/12 + (12.055/12)²+....+(12.055/12)^n-1)
A₃₀₀ = 80 000(12.055/12)³⁰⁰ - M(1+ 12.055/12 + (12.055/12)²+....+12.055/12²⁹⁹)
0 = 80 000(12.055/12)³⁰⁰ - M(1+ 12.055/12 + (12.055/12)²+....+12.055/12²⁹⁹)
M = 80 000(12.055/12)³⁰⁰ / (((12.055/12)³⁰⁰-1)/((12.055/12)-1))
=$491.27

Note I get 12.055/12 from 1+ 0.055/12

Answering for those that are looking for the answer in the future. If anyone needs clarification with this, let me know!

Sequence and series (1 Viewer)

fullonoob

fail engrish? unpossible!

fullonoob

fail engrish? unpossible!

Lukybear

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addikaye03

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fullonoob

fail engrish? unpossible!

Lukybear

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Lukybear

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fullonoob

fail engrish? unpossible!

fullonoob

fail engrish? unpossible!

fullonoob

fail engrish? unpossible!

gwc

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