<PRE><B>Question 1175621</B>
&lt;br&gt;
His investments for each year form an arithmetic sequence:&lt;br&gt;
2000, 2600, 3200, 3800, 4400, 5000, 5600, 6200, 6800, ...&lt;br&gt;
The fastest path to the answer is to add up the successive amounts until the total reaches 37000 or more.  You will quickly find that his total investment over 8 years is less than 37000 and the total over 9 years is greater then 37000.&lt;br&gt;
For an algebraic solution, using the formula for the sum of an arithmetic sequence....&lt;br&gt;
first payment: 2000
payment in n-th year: 2000+600(n-1) = 1400+600n&lt;br&gt;
Since the sequence is arithmetic, the sum of payments over n years is the number of years, multiplied by the average of the amounts from the first and last years:&lt;br&gt;
{{{S = n((2000+(1400+600n))/2) = n(1700+300n)}}}&lt;br&gt;
We want that sum to be 37000:&lt;br&gt;
{{{n(1700+300n) = 37000}}}
{{{1700n+300n^2 = 37000}}}
{{{300n^2+1700n-37000 = 0}}}
{{{3n^2+17n-370 = 0}}}&lt;br&gt;
That quadratic does not factor; you need to find the solution using the quadratic formula or a graphing calculator.  Of those two options, clearly the graphing calculator is more efficient.&lt;br&gt;
But you found the answer much faster by performing a few simple additions....&lt;br&gt;
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