Lesson How long will it take for an ordinary annuity to get an assigned value?
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<H2>How long will it take for an ordinary annuity to get an assigned value?</H2> <H3>Problem 1</H3>How long does $1000 have to be deposited into a savings account at the end of each month to accumulate to $36,000, if the annual interest is 6.4% compounded monthly? <B>Solution</B> <pre> Use the formula for an Ordinary Annuity saving account compounded monthly FV = {{{P*(((1+r/12)^n-1)/((r/12)))}}} where FV is the future value, P is the payment at the end of each month, r is the interest rate per year expressed as decimal, n is the number of monthly deposits (of months). So, we need to find " n " from this equation {{{((1+0.064/12)^n-1)/((0.064/12))}}} = {{{FV/P}}} = {{{36000/1000}}, which is the same as {{{((1+0.064/12)^n-1)/(0.064/12)}}} = 36, {{{(1+0.064/12)^n-1}}} = {{{(0.064/12)*36}}}. Rewrite it in this form {{{(1+0.064/12)^n-1}}} = 0.192, {{{(1+0.064/12)^n}}} = 1 + 0.192 = 1.192. Take logarithm base 10 of both sides n*log(1+0.064/12) = log(1.192) and calculate n = {{{log((1.192))/log((1+0.064/12))}}} = 33.01885 months. Round it to the closest greater month, which is 34 months, in order for the bank would be in position to compete the last compounding. 34 months is the same as 2 years and 10 months. <U>ANSWER</U>. 34 months, or 2 years and 10 months. </pre> <H3>Problem 2</H3>If money can earn 4.0% compounded monthly, how long will it take you to accumulate $1,000,000 in your saving plan if you contribute $1440 at the end of each month? <B>Solution</B> <pre> Use the formula for an Ordinary Annuity saving account compounded monthly FV = {{{P*(((1+r/12)^n-1)/((r/12)))}}} where FV is the future value, P is the payment at the end of each month, r is the interest rate per year expressed as decimal, n is the number of monthly deposits (of months). So, we need to find " n " from this equation {{{((1+0.04/12)^n-1)/((0.04/12))}}} = {{{FV/P}}} = {{{1000000/1440}}, which is the same as {{{((1+0.04/12)^n-1)/(0.04/12)}}} = 694.4444444 , {{{(1+0.04/12)^n-1}}} = {{{(0.04/12)*694.4444444}}}. Rewrite it in this form {{{(1+0.04/12)^n-1}}} = 2.314815, {{{(1+0.04/12)^n}}} = 1 + 2.314815 = 3.314815. Take logarithm base 10 of both sides n*log(1+0.04/12) = log(3.314815) and calculate n = {{{log((3.314815))/log((1+0.04/12))}}} = 360.1194131 months. Round it to the closest greater month, which is 361 months, in order for the bank would be in position to compete the last compounding. 361 months is the same as 30 years and 1 month. <U>ANSWER</U>. 361 months, or 30 years and 1 month. </pre> On ordinary annuity saving plan, see my lessons in this site - <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Ordinary-Annuity-saving-plans-and-geometric-progressions.lesson>Ordinary Annuity saving plans and geometric progressions</A> - <A HREF=https://www.algebra.com/algebra/homework/Sequences-and-series/Solved-problem-on-Ordinary-Annuity-saving-plans.lesson>Solved problems on Ordinary Annuity saving plans</A> These lessons contain EVERYTHING you need to know about this subject, in clear and compact form. When you learn from these lessons, you will be able to solve similar problems and to perform similar calculations in semi-automatic mode. 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