Final Payment Value Compound Interest Question
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A demand loan of $$4000.00$ is repaid by payments of $$2000.00$ after two years, $$2000.00$ after four years, and a final payment after six years.
Interest is $6%$ compounded quarterly for the first two years, $7%$ compounded annually for the next two years, and 7% compounded semi-annually thereafter. What is the size of the final payment?
I know how to make all these precursor calculations but I'm very lost on which formula to use to find the final amount.
payment. $3000(1+0.0175)^{0.5}$
Payment $3000(1+0.04)^{2}$
algebra-precalculus finance
edited Nov 19 at 5:30
NoChance
3,59621221
asked Nov 19 at 5:03
John Deer
111
|
show 3 more comments
up vote
1
down vote
favorite
A demand loan of $$4000.00$ is repaid by payments of $$2000.00$ after two years, $$2000.00$ after four years, and a final payment after six years.
Interest is $6%$ compounded quarterly for the first two years, $7%$ compounded annually for the next two years, and 7% compounded semi-annually thereafter. What is the size of the final payment?
I know how to make all these precursor calculations but I'm very lost on which formula to use to find the final amount.
payment. $3000(1+0.0175)^{0.5}$
Payment $3000(1+0.04)^{2}$
algebra-precalculus finance
edited Nov 19 at 5:30
NoChance
3,59621221
asked Nov 19 at 5:03
John Deer
111
1
Far better to use understanding than to use a formula.
– Gerry Myerson
Nov 19 at 5:08
I get that but I'm struggling with the concept and I think knowing the formula would allow me the context to mentally frame it better.
– John Deer
Nov 19 at 5:13
I think understanding the mathematics would be a better idea, but each to his/her own.
– Gerry Myerson
Nov 19 at 5:19
If you would like to try and give me a better understanding I would very much like to hear it.
– John Deer
Nov 19 at 5:24
1
OK. What's the balance after two years, before the first repayment? What's the balance after two years, after the first reapyment? Same question for after 4 years, before and after the second repayment. Then, what's the balance at the end of six years?
– Gerry Myerson
Nov 19 at 5:27
|
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A demand loan of $$4000.00$ is repaid by payments of $$2000.00$ after two years, $$2000.00$ after four years, and a final payment after six years.
Interest is $6%$ compounded quarterly for the first two years, $7%$ compounded annually for the next two years, and 7% compounded semi-annually thereafter. What is the size of the final payment?
I know how to make all these precursor calculations but I'm very lost on which formula to use to find the final amount.
payment. $3000(1+0.0175)^{0.5}$
Payment $3000(1+0.04)^{2}$
algebra-precalculus finance
edited Nov 19 at 5:30
NoChance
3,59621221
asked Nov 19 at 5:03
John Deer
111
A demand loan of $$4000.00$ is repaid by payments of $$2000.00$ after two years, $$2000.00$ after four years, and a final payment after six years.
Interest is $6%$ compounded quarterly for the first two years, $7%$ compounded annually for the next two years, and 7% compounded semi-annually thereafter. What is the size of the final payment?
I know how to make all these precursor calculations but I'm very lost on which formula to use to find the final amount.
payment. $3000(1+0.0175)^{0.5}$
Payment $3000(1+0.04)^{2}$
algebra-precalculus finance
algebra-precalculus finance
edited Nov 19 at 5:30
NoChance
3,59621221
asked Nov 19 at 5:03
John Deer
111
edited Nov 19 at 5:30
NoChance
3,59621221
asked Nov 19 at 5:03
John Deer
111
edited Nov 19 at 5:30
NoChance
3,59621221
edited Nov 19 at 5:30
NoChance
3,59621221
edited Nov 19 at 5:30
NoChance
3,59621221
3,59621221
asked Nov 19 at 5:03
John Deer
111
asked Nov 19 at 5:03
John Deer
111
asked Nov 19 at 5:03
John Deer
111
111
1
Far better to use understanding than to use a formula.
– Gerry Myerson
Nov 19 at 5:08
I get that but I'm struggling with the concept and I think knowing the formula would allow me the context to mentally frame it better.
– John Deer
Nov 19 at 5:13
I think understanding the mathematics would be a better idea, but each to his/her own.
– Gerry Myerson
Nov 19 at 5:19
If you would like to try and give me a better understanding I would very much like to hear it.
– John Deer
Nov 19 at 5:24
1
OK. What's the balance after two years, before the first repayment? What's the balance after two years, after the first reapyment? Same question for after 4 years, before and after the second repayment. Then, what's the balance at the end of six years?
– Gerry Myerson
Nov 19 at 5:27
|
show 3 more comments
1
Far better to use understanding than to use a formula.
– Gerry Myerson
Nov 19 at 5:08
I get that but I'm struggling with the concept and I think knowing the formula would allow me the context to mentally frame it better.
– John Deer
Nov 19 at 5:13
I think understanding the mathematics would be a better idea, but each to his/her own.
– Gerry Myerson
Nov 19 at 5:19
If you would like to try and give me a better understanding I would very much like to hear it.
– John Deer
Nov 19 at 5:24
1
OK. What's the balance after two years, before the first repayment? What's the balance after two years, after the first reapyment? Same question for after 4 years, before and after the second repayment. Then, what's the balance at the end of six years?
– Gerry Myerson
Nov 19 at 5:27
1
1
Far better to use understanding than to use a formula.
– Gerry Myerson
Nov 19 at 5:08
Far better to use understanding than to use a formula.
– Gerry Myerson
Nov 19 at 5:08
I get that but I'm struggling with the concept and I think knowing the formula would allow me the context to mentally frame it better.
– John Deer
Nov 19 at 5:13
I get that but I'm struggling with the concept and I think knowing the formula would allow me the context to mentally frame it better.
– John Deer
Nov 19 at 5:13
I think understanding the mathematics would be a better idea, but each to his/her own.
– Gerry Myerson
Nov 19 at 5:19
I think understanding the mathematics would be a better idea, but each to his/her own.
– Gerry Myerson
Nov 19 at 5:19
If you would like to try and give me a better understanding I would very much like to hear it.
– John Deer
Nov 19 at 5:24
If you would like to try and give me a better understanding I would very much like to hear it.
– John Deer
Nov 19 at 5:24
1
1
OK. What's the balance after two years, before the first repayment? What's the balance after two years, after the first reapyment? Same question for after 4 years, before and after the second repayment. Then, what's the balance at the end of six years?
– Gerry Myerson
Nov 19 at 5:27
OK. What's the balance after two years, before the first repayment? What's the balance after two years, after the first reapyment? Same question for after 4 years, before and after the second repayment. Then, what's the balance at the end of six years?
– Gerry Myerson
Nov 19 at 5:27
|
show 3 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
The final payment can be calculated in steps by figuring how much the loan has grown with interest over each time period before each of the stage payments. I'm assuming the interest for each time period is simply the annual rate divided by the number of time periods per year.
$$P_F = ((4000cdot 1.015^8 -2000)1.07^2 - 2000)1.035^4 = $997.30$$
answered Nov 19 at 5:50
Phil H
3,9302312
I'm confused where you're getting your N value as I thought it was numbers of years divided by interests compounds per year so for the first 2 years wouldn't it be 4000(1.015)^0.5 as it Would be 2 years divided by 4 (quarterly interest compounds)
– John Deer
Nov 19 at 5:57
No, it's 1.015 to the power of the number of quarters in 2 years which is 8.Then 1.07 to the power of the number of years which is 2 and finally 1.035 to the power of the number of half years in 2 years which is 4.
– Phil H
Nov 19 at 7:18
add a comment |
up vote
0
down vote
$L=4000$, $p_2=2000$, $p_4=2000$, $i^{(4)} =6%$, $i^{(1)} =7%$, $i^{(2)} =7%$. Find $p_6$ solving
$$
L=frac{p_2}{left(1+frac{i^{(4)}}{4}right)^{2times 4}} +frac{p_4}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}}+frac{p_6}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}timesleft(1+frac{i^{(2)}}{2}right)^{2times 2}}
$$
answered Dec 2 at 0:51
alexjo
12.2k1329
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The final payment can be calculated in steps by figuring how much the loan has grown with interest over each time period before each of the stage payments. I'm assuming the interest for each time period is simply the annual rate divided by the number of time periods per year.
$$P_F = ((4000cdot 1.015^8 -2000)1.07^2 - 2000)1.035^4 = $997.30$$
answered Nov 19 at 5:50
Phil H
3,9302312
I'm confused where you're getting your N value as I thought it was numbers of years divided by interests compounds per year so for the first 2 years wouldn't it be 4000(1.015)^0.5 as it Would be 2 years divided by 4 (quarterly interest compounds)
– John Deer
Nov 19 at 5:57
No, it's 1.015 to the power of the number of quarters in 2 years which is 8.Then 1.07 to the power of the number of years which is 2 and finally 1.035 to the power of the number of half years in 2 years which is 4.
– Phil H
Nov 19 at 7:18
add a comment |
up vote
1
down vote
The final payment can be calculated in steps by figuring how much the loan has grown with interest over each time period before each of the stage payments. I'm assuming the interest for each time period is simply the annual rate divided by the number of time periods per year.
$$P_F = ((4000cdot 1.015^8 -2000)1.07^2 - 2000)1.035^4 = $997.30$$
answered Nov 19 at 5:50
Phil H
3,9302312
I'm confused where you're getting your N value as I thought it was numbers of years divided by interests compounds per year so for the first 2 years wouldn't it be 4000(1.015)^0.5 as it Would be 2 years divided by 4 (quarterly interest compounds)
– John Deer
Nov 19 at 5:57
No, it's 1.015 to the power of the number of quarters in 2 years which is 8.Then 1.07 to the power of the number of years which is 2 and finally 1.035 to the power of the number of half years in 2 years which is 4.
– Phil H
Nov 19 at 7:18
add a comment |
up vote
1
down vote
up vote
1
down vote
The final payment can be calculated in steps by figuring how much the loan has grown with interest over each time period before each of the stage payments. I'm assuming the interest for each time period is simply the annual rate divided by the number of time periods per year.
$$P_F = ((4000cdot 1.015^8 -2000)1.07^2 - 2000)1.035^4 = $997.30$$
answered Nov 19 at 5:50
Phil H
3,9302312
The final payment can be calculated in steps by figuring how much the loan has grown with interest over each time period before each of the stage payments. I'm assuming the interest for each time period is simply the annual rate divided by the number of time periods per year.
$$P_F = ((4000cdot 1.015^8 -2000)1.07^2 - 2000)1.035^4 = $997.30$$
answered Nov 19 at 5:50
Phil H
3,9302312
answered Nov 19 at 5:50
Phil H
3,9302312
answered Nov 19 at 5:50
Phil H
3,9302312
answered Nov 19 at 5:50
Phil H
3,9302312
3,9302312
I'm confused where you're getting your N value as I thought it was numbers of years divided by interests compounds per year so for the first 2 years wouldn't it be 4000(1.015)^0.5 as it Would be 2 years divided by 4 (quarterly interest compounds)
– John Deer
Nov 19 at 5:57
No, it's 1.015 to the power of the number of quarters in 2 years which is 8.Then 1.07 to the power of the number of years which is 2 and finally 1.035 to the power of the number of half years in 2 years which is 4.
– Phil H
Nov 19 at 7:18
add a comment |
I'm confused where you're getting your N value as I thought it was numbers of years divided by interests compounds per year so for the first 2 years wouldn't it be 4000(1.015)^0.5 as it Would be 2 years divided by 4 (quarterly interest compounds)
– John Deer
Nov 19 at 5:57
No, it's 1.015 to the power of the number of quarters in 2 years which is 8.Then 1.07 to the power of the number of years which is 2 and finally 1.035 to the power of the number of half years in 2 years which is 4.
– Phil H
Nov 19 at 7:18
I'm confused where you're getting your N value as I thought it was numbers of years divided by interests compounds per year so for the first 2 years wouldn't it be 4000(1.015)^0.5 as it Would be 2 years divided by 4 (quarterly interest compounds)
– John Deer
Nov 19 at 5:57
I'm confused where you're getting your N value as I thought it was numbers of years divided by interests compounds per year so for the first 2 years wouldn't it be 4000(1.015)^0.5 as it Would be 2 years divided by 4 (quarterly interest compounds)
– John Deer
Nov 19 at 5:57
No, it's 1.015 to the power of the number of quarters in 2 years which is 8.Then 1.07 to the power of the number of years which is 2 and finally 1.035 to the power of the number of half years in 2 years which is 4.
– Phil H
Nov 19 at 7:18
No, it's 1.015 to the power of the number of quarters in 2 years which is 8.Then 1.07 to the power of the number of years which is 2 and finally 1.035 to the power of the number of half years in 2 years which is 4.
– Phil H
Nov 19 at 7:18
add a comment |
up vote
0
down vote
$L=4000$, $p_2=2000$, $p_4=2000$, $i^{(4)} =6%$, $i^{(1)} =7%$, $i^{(2)} =7%$. Find $p_6$ solving
$$
L=frac{p_2}{left(1+frac{i^{(4)}}{4}right)^{2times 4}} +frac{p_4}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}}+frac{p_6}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}timesleft(1+frac{i^{(2)}}{2}right)^{2times 2}}
$$
answered Dec 2 at 0:51
alexjo
12.2k1329
add a comment |
up vote
0
down vote
$L=4000$, $p_2=2000$, $p_4=2000$, $i^{(4)} =6%$, $i^{(1)} =7%$, $i^{(2)} =7%$. Find $p_6$ solving
$$
L=frac{p_2}{left(1+frac{i^{(4)}}{4}right)^{2times 4}} +frac{p_4}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}}+frac{p_6}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}timesleft(1+frac{i^{(2)}}{2}right)^{2times 2}}
$$
answered Dec 2 at 0:51
alexjo
12.2k1329
add a comment |
up vote
0
down vote
up vote
0
down vote
$L=4000$, $p_2=2000$, $p_4=2000$, $i^{(4)} =6%$, $i^{(1)} =7%$, $i^{(2)} =7%$. Find $p_6$ solving
$$
L=frac{p_2}{left(1+frac{i^{(4)}}{4}right)^{2times 4}} +frac{p_4}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}}+frac{p_6}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}timesleft(1+frac{i^{(2)}}{2}right)^{2times 2}}
$$
answered Dec 2 at 0:51
alexjo
12.2k1329
$L=4000$, $p_2=2000$, $p_4=2000$, $i^{(4)} =6%$, $i^{(1)} =7%$, $i^{(2)} =7%$. Find $p_6$ solving
$$
L=frac{p_2}{left(1+frac{i^{(4)}}{4}right)^{2times 4}} +frac{p_4}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}}+frac{p_6}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}timesleft(1+frac{i^{(2)}}{2}right)^{2times 2}}
$$
answered Dec 2 at 0:51
alexjo
12.2k1329
answered Dec 2 at 0:51
alexjo
12.2k1329
answered Dec 2 at 0:51
alexjo
12.2k1329
answered Dec 2 at 0:51
alexjo
12.2k1329
12.2k1329
add a comment |
add a comment |
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1
Far better to use understanding than to use a formula.
– Gerry Myerson
Nov 19 at 5:08
I get that but I'm struggling with the concept and I think knowing the formula would allow me the context to mentally frame it better.
– John Deer
Nov 19 at 5:13
I think understanding the mathematics would be a better idea, but each to his/her own.
– Gerry Myerson
Nov 19 at 5:19
If you would like to try and give me a better understanding I would very much like to hear it.
– John Deer
Nov 19 at 5:24
1
OK. What's the balance after two years, before the first repayment? What's the balance after two years, after the first reapyment? Same question for after 4 years, before and after the second repayment. Then, what's the balance at the end of six years?
– Gerry Myerson
Nov 19 at 5:27