Question 1153297
.


It works by the same way, as if the account is compounded quarterly.



<pre>
It is a classic Ordinary Annuity saving plan. The general formula is 


    FV = {{{P*(((1+r)^n-1)/r)}}},    (1)


where  FV is the future value of the account;  P is the quarterly deposit; r is the quarterly percentage yield presented as a decimal; 
n is the number of deposits (= the number of quarters, in this case).


Under the given conditions, P = 100;  r = 0.06/4.  So, according to the formula (1), you get at the end of the n-th quarter


    FV = {{{100*(((1+0.06/4)^n-1)/((0.06/4)))}}} = {{{100*((1.015^n-1)/0.015)}}}.



You want to find n from the inequality

    {{{100*((1.015^n-1)/0.015)}}} >= 20000.


Then

    {{{1.015^n-1}}} >= {{{(20000/100)*0.015}}} = 3

    {{{1.015^n}}} >= 4

    n*log(1.015) >= log(4)

    n >= {{{log(4)/log(1.015)}}} = 93.11


The <U>preliminary ANSWER</U> is  94 quarters, or  23 years and 2 quarters.


Let's check it. After 93 quarters, the amount will be

    FV = {{{100*(((1+0.06/4)^93-1)/((0.06/4)))}}} = 19955.95.


So, actually  93 quarters and 1 month is enough to get $20000.


<U>The final answer</U>.  93 quarters and 1 month, or  23 years and 4 months.
</pre>

Solved.


-----------------


On Ordinary Annuity saving plans, &nbsp;see the lessons

&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>

in this site.


The lessons contain &nbsp;EVERYTHING &nbsp;you need to know about this subject, &nbsp;in clear and compact form.


When you learn from these lessons, &nbsp;you will be able to do similar calculations in semi-automatic mode.