document.write( "Question 1194508: How much must be deposited at the end of each month for 3.5 years to accumulate to $1,801.00 at 8%
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Algebra.Com's Answer #826725 by math_tutor2020(3817)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "1 year = 12 months \n" ); document.write( "3.5*1 year = 3.5*12 months \n" ); document.write( "3.5 years = 42 months\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "We want the future value to be $1,801 and have it occur at the 42 month mark. \n" ); document.write( "The annual interest rate in decimal form is r = 0.08\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Divide by 12 to get the monthly version: r/12 = 0.08/12 = 0.00666666666667 \n" ); document.write( "which is approximate.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The future value annuity formula to use is this \n" ); document.write( "FV = P*( (1+i)^n - 1)/i \n" ); document.write( "This is an ordinary annuity and not \"annuity due\". \n" ); document.write( "This is because the deposits happen at the end of each month, rather at the beginning.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "In this case, \n" ); document.write( "FV = future value = 1801 \n" ); document.write( "P = unknown monthly deposit \n" ); document.write( "i = 0.00666666666667 approximately \n" ); document.write( "n = 42 months\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let's solve for P \n" ); document.write( "FV = P*( (1+i)^n - 1)/i \n" ); document.write( "1801 = P*( (1+0.00666666666667)^42 - 1)/0.00666666666667 \n" ); document.write( "1801 = P*48.2851385181578 \n" ); document.write( "P = 1801/48.2851385181578 \n" ); document.write( "P = 37.2992613311594 \n" ); document.write( "P = 37.30\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Answer: $37.30 \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |