Question 1209569
<br>
For an arithmetic sequence....<br>
Sum = (number of terms)*(average of terms)<br>
Number of terms: ((last term minus first term)/common difference)+1<br>
={{{(x-1)/3+1=(x+2)/3}}}<br>
Average of terms = average of first and last terms = {{{(x+1)/2}}}<br>
The sum is 287:<br>
{{{((x+2)/3)((x+1)/2)=287}}}<br>
{{{(x^2+3x+2)/6=287}}}<br>
{{{x^2+3x+2=1722}}}<br>
{{{x^2+3x-1720=0}}}<br>
{{{(x+43)(x-40)=0}}}<br>
-----------------------------------------------------------------------<br>
A hint for doing that factoring.... We want to write<br>
{{{x^2+3x-1720=(x+a)(x-b)}}}<br>
Since the linear coefficient is 3, a and b are "close together".  So let them be "approximately equal".  Then 1720 is "close to" a perfect square. The square root of 1720 is a bit more than 40; and since the product ab has final digit 0, one of a or b should be 40.  Then a bit of arithmetic shows that a and b are in fact 43 and 40.<br>
-----------------------------------------------------------------------<br>
Obviously ignore the negative solution....<br>
ANSWER: x=40<br>