Future value

Suppose you invested $1000 per quarter over a 15 years period. If money earns an anual rate of 6.5% compounded quarterly, how much would be available at the end of the time period? How much is the interest earn?

Correct result:

f =  100336.68 USD
i =  40336.68 USD

Solution:

$$\begin{gathered} r=4 \\ a=1000 \\ p=6.5\mathrm{\%}={\displaystyle \frac{6.5}{100}}=0.065 \\ n=15\cdot r=15\cdot 4=60 \\ q=p\mathrm{/}r=0.065\mathrm{/}4={\displaystyle \frac{13}{800}}\doteq 0.0163 \\ k=1+q=1+0.0163={\displaystyle \frac{813}{800}}\doteq 1.0163 \\ {f}_1=a=1000 \\ {f}_2=[k\cdot f_1+a]=[1.0163\cdot 1000+1000]={\displaystyle \frac{8065}{4}}=2016.25 \\ {f}_3=[k\cdot f_2+a]=[1.0163\cdot 2016.25+1000]=3049.01 \\ {f}_4=[k\cdot f_3+a]=[1.0163\cdot 3049.01+1000]=4098.56 \\ {f}_5=[k\cdot f_4+a]=[1.0163\cdot 4098.56+1000]=5165.16 \\ {f}_6=[k\cdot f_5+a]=[1.0163\cdot 5165.16+1000]=6249.09 \\ {f}_7=[k\cdot f_6+a]=[1.0163\cdot 6249.09+1000]=7350.64 \\ {f}_8=[k\cdot f_7+a]=[1.0163\cdot 7350.64+1000]=8470.09 \\ {f}_9=[k\cdot f_8+a]=[1.0163\cdot 8470.09+1000]=9607.73 \\ {f}_{10}=[k\cdot f_9+a]=[1.0163\cdot 9607.73+1000]=10763.86 \\ {f}_{11}=[k\cdot f_{10}+a]=[1.0163\cdot 10763.86+1000]=11938.77 \\ {f}_{12}=[k\cdot f_{11}+a]=[1.0163\cdot 11938.77+1000]=13132.78 \\ {f}_{13}=[k\cdot f_{12}+a]=[1.0163\cdot 13132.78+1000]=14346.19 \\ {f}_{14}=[k\cdot f_{13}+a]=[1.0163\cdot 14346.19+1000]=15579.32 \\ {f}_{15}=[k\cdot f_{14}+a]=[1.0163\cdot 15579.32+1000]=16832.48 \\ {f}_{16}=[k\cdot f_{15}+a]=[1.0163\cdot 16832.48+1000]=18106.01 \\ {f}_{17}=[k\cdot f_{16}+a]=[1.0163\cdot 18106.01+1000]=19400.23 \\ {f}_{18}=[k\cdot f_{17}+a]=[1.0163\cdot 19400.23+1000]=20715.48 \\ {f}_{19}=[k\cdot f_{18}+a]=[1.0163\cdot 20715.48+1000]=22052.11 \\ {f}_{20}=[k\cdot f_{19}+a]=[1.0163\cdot 22052.11+1000]=23410.46 \\ {f}_{21}=[k\cdot f_{20}+a]=[1.0163\cdot 23410.46+1000]=24790.88 \\ {f}_{22}=[k\cdot f_{21}+a]=[1.0163\cdot 24790.88+1000]=26193.73 \\ {f}_{23}=[k\cdot f_{22}+a]=[1.0163\cdot 26193.73+1000]=27619.38 \\ {f}_{24}=[k\cdot f_{23}+a]=[1.0163\cdot 27619.38+1000]=29068.19 \\ {f}_{25}=[k\cdot f_{24}+a]=[1.0163\cdot 29068.19+1000]=30540.55 \\ {f}_{26}=[k\cdot f_{25}+a]=[1.0163\cdot 30540.55+1000]=32036.83 \\ {f}_{27}=[k\cdot f_{26}+a]=[1.0163\cdot 32036.83+1000]=33557.43 \\ {f}_{28}=[k\cdot f_{27}+a]=[1.0163\cdot 33557.43+1000]=35102.74 \\ {f}_{29}=[k\cdot f_{28}+a]=[1.0163\cdot 35102.74+1000]=36673.16 \\ {f}_{30}=[k\cdot f_{29}+a]=[1.0163\cdot 36673.16+1000]=38269.1 \\ {f}_{31}=[k\cdot f_{30}+a]=[1.0163\cdot 38269.1+1000]=39890.97 \\ {f}_{32}=[k\cdot f_{31}+a]=[1.0163\cdot 39890.97+1000]={\displaystyle \frac{207696}{5}}=41539.2 \\ {f}_{33}=[k\cdot f_{32}+a]=[1.0163\cdot 41539.2+1000]=43214.21 \\ {f}_{34}=[k\cdot f_{33}+a]=[1.0163\cdot 43214.21+1000]=44916.44 \\ {f}_{35}=[k\cdot f_{34}+a]=[1.0163\cdot 44916.44+1000]=46646.33 \\ {f}_{36}=[k\cdot f_{35}+a]=[1.0163\cdot 46646.33+1000]=48404.33 \\ {f}_{37}=[k\cdot f_{36}+a]=[1.0163\cdot 48404.33+1000]=50190.9 \\ {f}_{38}=[k\cdot f_{37}+a]=[1.0163\cdot 50190.9+1000]={\displaystyle \frac{104013}{2}}=52006.5 \\ {f}_{39}=[k\cdot f_{38}+a]=[1.0163\cdot 52006.5+1000]=53851.61 \\ {f}_{40}=[k\cdot f_{39}+a]=[1.0163\cdot 53851.61+1000]=55726.7 \\ {f}_{41}=[k\cdot f_{40}+a]=[1.0163\cdot 55726.7+1000]=57632.26 \\ {f}_{42}=[k\cdot f_{41}+a]=[1.0163\cdot 57632.26+1000]=59568.78 \\ {f}_{43}=[k\cdot f_{42}+a]=[1.0163\cdot 59568.78+1000]=61536.77 \\ {f}_{44}=[k\cdot f_{43}+a]=[1.0163\cdot 61536.77+1000]=63536.74 \\ {f}_{45}=[k\cdot f_{44}+a]=[1.0163\cdot 63536.74+1000]=65569.21 \\ {f}_{46}=[k\cdot f_{45}+a]=[1.0163\cdot 65569.21+1000]=67634.71 \\ {f}_{47}=[k\cdot f_{46}+a]=[1.0163\cdot 67634.71+1000]=69733.77 \\ {f}_{48}=[k\cdot f_{47}+a]=[1.0163\cdot 69733.77+1000]=71866.94 \\ {f}_{49}=[k\cdot f_{48}+a]=[1.0163\cdot 71866.94+1000]=74034.78 \\ {f}_{50}=[k\cdot f_{49}+a]=[1.0163\cdot 74034.78+1000]=76237.85 \\ {f}_{51}=[k\cdot f_{50}+a]=[1.0163\cdot 76237.85+1000]=78476.72 \\ {f}_{52}=[k\cdot f_{51}+a]=[1.0163\cdot 78476.72+1000]=80751.97 \\ {f}_{53}=[k\cdot f_{52}+a]=[1.0163\cdot 80751.97+1000]=83064.19 \\ {f}_{54}=[k\cdot f_{53}+a]=[1.0163\cdot 83064.19+1000]=85413.98 \\ {f}_{55}=[k\cdot f_{54}+a]=[1.0163\cdot 85413.98+1000]=87801.96 \\ {f}_{56}=[k\cdot f_{55}+a]=[1.0163\cdot 87801.96+1000]=90228.74 \\ {f}_{57}=[k\cdot f_{56}+a]=[1.0163\cdot 90228.74+1000]=92694.96 \\ {f}_{58}=[k\cdot f_{57}+a]=[1.0163\cdot 92694.96+1000]={\displaystyle \frac{380805}{4}}=95201.25 \\ {f}_{59}=[k\cdot f_{58}+a]=[1.0163\cdot 95201.25+1000]=97748.27 \\ {f}_{60}=[k\cdot f_{59}+a]=[1.0163\cdot 97748.27+1000]=100336.68 \\ f=f_{60}=100336.68=100336.68\text{ USD} \end{gathered}$$

$i=f-n\cdot a=100336.68-60\cdot 1000=40336.68\text{ USD}$

Try another example

Showing 0 comments:

Next similar math problems:

• Present value
  A bank loans a family $90,000 at 4.5% annual interest rate to purchase a house. The family agrees to pay the loan off by making monthly payments over a 15 year period. How much should the monthly payment be in order to pay off the debt in 15 years?
• Compound Interest
  If you deposit $6000 in an account paying 6.5% annual interest compounded quarterly, how long until there is $12600 in the account?
• Retirement annuity
  How much will it cost to purchase a two-level retirement annuity that will pay $2000 at the end of every month for the first 10 years, and $3000 per month for the next 15 years? Assume that the payment represent a rate of return to the person receiving th
• Savings
  Suppose on your 21st birthday you begin making monthly payments of $500 into an account that pays 8% compounded monthly. If you continue the payments untill your 51st birthday (30 years), How much money will be in your account? How much of it is interest?
• Interest
  Calculate how much you earn for 10 years 43000 deposit if the interest rate is 1.3% and the interest period is a quarter.
• Saving 9
  An amount if $ 2000 is invested at an interest of 5% per month. if $ 200 is added at the beginning of each successive month but no withdrawals. Give an expression for the value accumulated after n months. After how many months will the amount have accumul
• Investment
  1000$ is invested at 10% compound interest. What factor is the capital multiplied by each year? How much will be there after n=12 years?
• If you 3
  If you deposit $4500 at 5% annual interest compound quarterly, how much money will be in the account after 10 years?
• How much 2
  How much money would you need to deposit today at 5% annual interest compounded monthly to have $2000 in the account after 9 years?
• If you 4
  If you deposit $2500 in an account paying 11% annual interest compounded quarterly, how long until there is the $4500 in the account?
• Loan
  Apply for a $ 59000 loan, the loan repayment period is 8 years, the interest rate 7%. How much should I pay for every month (or every year if paid yearly). Example is for practise geometric progression and/or periodic payment for an annuity.
• If you 2
  If you deposit $4000 into an account paying 9% annual interest compounded monthly, how long until there is $10000 in the account?
• Investment 2
  Jack invested $5000 in a 5-month term deposit at 4.7% p. A. . At the end of the 5-months, jack reinvested the maturity value from the first deposit into an 11-month term deposit at 7.3% p. A. What is the maturity value at the end of the second term deposi
• Bank
  Paul put 10000 in the bank for 6 years. Calculate how much you will have in the bank if he not pick earned interest or change deposit conditions. The annual interest rate is 3.5%, and the tax on interest is 10%.
• Compound interest 3
  After 8 years, what is the total amount of a compound interest investment of $25,000 at 3% interest, compounded quarterly? (interest is now dream - in the year 2019)
• Semiannually compound interest
  If you deposit $5000 into an account paying 8.25% annual interest compounded semiannually, how long until there is $9350 in the account?
• Savings
  The depositor regularly wants to invest the same amount of money in the financial institution at the beginning of the year and wants to save 10,000 euros at the end of the tenth year. What amount should he deposit if the annual interest rate for the annua