Question 1207351
.
How long does $1000 have to be deposited into a savings account at the end of each month 
to accumulate to $36,000 if interest is 6.4% compounded monthly?
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<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. 
34 months is the same as 2 years and 10 months.


<U>ANSWER</U>.  34 months, or 2 years and 10 months.
</pre>

Solved.


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On ordinary annuity saving plan, &nbsp;see my lessons in this site 


&nbsp;&nbsp;&nbsp;&nbsp;- <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>

&nbsp;&nbsp;&nbsp;&nbsp;- <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>