<PRE><B>Question 1153229</B>
&lt;br&gt;
The numbers of seats form an arithmetic sequence:&lt;br&gt;
12, 15, 18, ..., 12+3(n-1)&lt;br&gt;
We need to determine the number of rows (n, the number of terms in the sequence) for which the sum is 255.  Algebraically, the sum is&lt;br&gt;
(number of terms) times (average of all the terms)&lt;br&gt;
which for an arithmetic sequence is&lt;br&gt;
(number of terms) times (average of first and last terms)&lt;br&gt;
{{{255 = (n)((12+(12+3(n-1)))/2)}}}
{{{510 = n(3n+21)}}}
{{{510 = 3n^2+21n}}}
{{{3n^2+21n-510 = 0}}}
{{{n^2+7n-170 = 0}}}
{{{(n+17)(n-10)= 0}}}&lt;br&gt;
n = -17  or n = 10; but negative numbers don&#39;t make sense for the problem.  So n=10.&lt;br&gt;
ANSWER: 10 rows&lt;br&gt;
If a formal algebraic solution is not required, the problem can be solved fairly easily using mental arithmetic.&lt;br&gt;
The sum of 255 is the product of two numbers -- the number of rows, and the average number of seats in each row.  That means that twice the sum, 510, is the product of two whole numbers -- the number of rows, and the SUM of the numbers in the first and last rows.&lt;br&gt;
Knowing that the sum of the numbers of seats in the first and last rows is a multiple of 3 (because the number in each row is a multiple of 3), some quick mental arithmetic can find the solution.&lt;br&gt;
Perhaps the most obvious pair of numbers to try is&lt;br&gt;
510 = 10*51&lt;br&gt;
This potential solution requires 10 rows, with 12+9(3) = 39 seats in the last row, making the sum of the numbers of seats in the first and last rows 12+39 = 51 -- and that satisfies the conditions of the problem.&lt;br&gt;
So again the answer is 10 rows.&lt;br&gt;
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