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}=1.0\cdot 10^5\text{ USD} \end{gathered}$$

$i=f-n\cdot a=100336.68-60\cdot 1000=40336.68\text{ USD}=4.0\cdot 10^4\text{ USD}$

Try another example

Showing 0 comments:

Related math problems and questions:

• 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?
• Suppose 3
  Suppose that a couple invested Php 50 000 in an account when their child was born, to prepare for the child's college education. If the average interest rate is 4.4% compounded annually, a, Give an exponential model for the situation b, Will the money be
• 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?
• You take
  You take out Php 20 000 loan at 5% interest rate. If the interest is compounded annually, a. Give an exponential model for the situation b. How much Will you owe after 10 years?
• 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)