Final Payment Value Compound Interest Question












1














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.




  1. payment. $3000(1+0.0175)^{0.5}$


  2. Payment $3000(1+0.04)^{2}$







algebra-precalculus finance







share|cite|improve this question




















  • 1




    Far better to use understanding than to use a formula.
    – Gerry Myerson
    Nov 19 '18 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 '18 at 5:13










  • I think understanding the mathematics would be a better idea, but each to his/her own.
    – Gerry Myerson
    Nov 19 '18 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 '18 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 '18 at 5:27
















1














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.




  1. payment. $3000(1+0.0175)^{0.5}$


  2. Payment $3000(1+0.04)^{2}$







algebra-precalculus finance







share|cite|improve this question




















  • 1




    Far better to use understanding than to use a formula.
    – Gerry Myerson
    Nov 19 '18 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 '18 at 5:13










  • I think understanding the mathematics would be a better idea, but each to his/her own.
    – Gerry Myerson
    Nov 19 '18 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 '18 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 '18 at 5:27














1












1








1







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.




  1. payment. $3000(1+0.0175)^{0.5}$


  2. Payment $3000(1+0.04)^{2}$







algebra-precalculus finance







share|cite|improve this question















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.




  1. payment. $3000(1+0.0175)^{0.5}$


  2. Payment $3000(1+0.04)^{2}$







algebra-precalculus finance




algebra-precalculus finance






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 19 '18 at 5:30









NoChance

3,62121221




3,62121221










asked Nov 19 '18 at 5:03









John Deer

111




111








  • 1




    Far better to use understanding than to use a formula.
    – Gerry Myerson
    Nov 19 '18 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 '18 at 5:13










  • I think understanding the mathematics would be a better idea, but each to his/her own.
    – Gerry Myerson
    Nov 19 '18 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 '18 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 '18 at 5:27














  • 1




    Far better to use understanding than to use a formula.
    – Gerry Myerson
    Nov 19 '18 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 '18 at 5:13










  • I think understanding the mathematics would be a better idea, but each to his/her own.
    – Gerry Myerson
    Nov 19 '18 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 '18 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 '18 at 5:27








1




1




Far better to use understanding than to use a formula.
– Gerry Myerson
Nov 19 '18 at 5:08




Far better to use understanding than to use a formula.
– Gerry Myerson
Nov 19 '18 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 '18 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 '18 at 5:13












I think understanding the mathematics would be a better idea, but each to his/her own.
– Gerry Myerson
Nov 19 '18 at 5:19




I think understanding the mathematics would be a better idea, but each to his/her own.
– Gerry Myerson
Nov 19 '18 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 '18 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 '18 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 '18 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 '18 at 5:27










2 Answers
2






active

oldest

votes


















1














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.



$$P_F = ((4000cdot 1.015^8 -2000)1.07^2 - 2000)1.035^4 = $997.30$$






share|cite|improve this answer





















  • 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 '18 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 '18 at 7:18



















0














$L=4000$, $p_2=2000$, $p_4=2000$, $i^{(4)} =6%$, $i^{(1)} =7%$, $i^{(2)} =7%$. Find $p_6$ solving



$$
L=frac{p_2}{left(1+frac{i^{(4)}}{4}right)^{2times 4}} +frac{p_4}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}}+frac{p_6}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}timesleft(1+frac{i^{(2)}}{2}right)^{2times 2}}
$$






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    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.



    $$P_F = ((4000cdot 1.015^8 -2000)1.07^2 - 2000)1.035^4 = $997.30$$






    share|cite|improve this answer





















    • 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 '18 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 '18 at 7:18
















    1














    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.



    $$P_F = ((4000cdot 1.015^8 -2000)1.07^2 - 2000)1.035^4 = $997.30$$






    share|cite|improve this answer





















    • 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 '18 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 '18 at 7:18














    1












    1








    1






    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.



    $$P_F = ((4000cdot 1.015^8 -2000)1.07^2 - 2000)1.035^4 = $997.30$$






    share|cite|improve this answer












    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.



    $$P_F = ((4000cdot 1.015^8 -2000)1.07^2 - 2000)1.035^4 = $997.30$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 19 '18 at 5:50









    Phil H

    4,0362312




    4,0362312












    • 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 7:18











    0














    $L=4000$, $p_2=2000$, $p_4=2000$, $i^{(4)} =6%$, $i^{(1)} =7%$, $i^{(2)} =7%$. Find $p_6$ solving



    $$
    L=frac{p_2}{left(1+frac{i^{(4)}}{4}right)^{2times 4}} +frac{p_4}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}}+frac{p_6}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}timesleft(1+frac{i^{(2)}}{2}right)^{2times 2}}
    $$






    share|cite|improve this answer


























      0














      $L=4000$, $p_2=2000$, $p_4=2000$, $i^{(4)} =6%$, $i^{(1)} =7%$, $i^{(2)} =7%$. Find $p_6$ solving



      $$
      L=frac{p_2}{left(1+frac{i^{(4)}}{4}right)^{2times 4}} +frac{p_4}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}}+frac{p_6}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}timesleft(1+frac{i^{(2)}}{2}right)^{2times 2}}
      $$






      share|cite|improve this answer
























        0












        0








        0






        $L=4000$, $p_2=2000$, $p_4=2000$, $i^{(4)} =6%$, $i^{(1)} =7%$, $i^{(2)} =7%$. Find $p_6$ solving



        $$
        L=frac{p_2}{left(1+frac{i^{(4)}}{4}right)^{2times 4}} +frac{p_4}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}}+frac{p_6}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}timesleft(1+frac{i^{(2)}}{2}right)^{2times 2}}
        $$






        share|cite|improve this answer












        $L=4000$, $p_2=2000$, $p_4=2000$, $i^{(4)} =6%$, $i^{(1)} =7%$, $i^{(2)} =7%$. Find $p_6$ solving



        $$
        L=frac{p_2}{left(1+frac{i^{(4)}}{4}right)^{2times 4}} +frac{p_4}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}}+frac{p_6}{left(1+frac{i^{(4)}}{4}right)^{2times 4}timesleft(1+frac{i^{(1)}}{1}right)^{2times 1}timesleft(1+frac{i^{(2)}}{2}right)^{2times 2}}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 0:51









        alexjo

        12.2k1329




        12.2k1329






























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