document.write( "Question 507557: Consider the arithmetic series 7+19+31+43+55+.....
\n" ); document.write( "What is the maximum number of terms for a sum of the series less than 1000?
\n" ); document.write( "What is the maximum number of terms for a sum of the series less than 2000?\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #340631 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
\r\n" );
document.write( "Sn = \"n%2F2\"(2a1 + (n-1)d) < 1000\r\n" );
document.write( "\r\n" );
document.write( "Sn = \"n%2F2\"[2*7 + (n-1)12] < 1000\r\n" );
document.write( "     \r\n" );
document.write( "Multiply both sides by 2\r\n" );
document.write( "\r\n" );
document.write( "      n[2*7 + (n-1)12] < 2000\r\n" );
document.write( "\r\n" );
document.write( "      n[14 + 12n - 12] < 2000             \r\n" );
document.write( "      \r\n" );
document.write( "            n[2 + 12n] < 2000\r\n" );
document.write( "\r\n" );
document.write( "             2n + 12n² < 2000\r\n" );
document.write( "\r\n" );
document.write( "      12n² + 2n - 2000 < 0\r\n" );
document.write( "\r\n" );
document.write( "Divide through by 2\r\n" );
document.write( "\r\n" );
document.write( "        6n² + n - 1000 < 0    \r\n" );
document.write( "\r\n" );
document.write( "Use the quadratic formula to determine the zeros\r\n" );
document.write( "of \r\n" );
document.write( "\r\n" );
document.write( "f(n) = 6n² + n - 1000\r\n" );
document.write( "\r\n" );
document.write( "They are approximately 12.82688011 and -12.99354678\r\n" );
document.write( "\r\n" );
document.write( "We know n cannot be a negative number.  We think therefore\r\n" );
document.write( "the answer will be the greatest integer less than 12.82688011\r\n" );
document.write( "which is 12.\r\n" );
document.write( "\r\n" );
document.write( "To prove it we substitute n = 12\r\n" );
document.write( "\r\n" );
document.write( "Sn = \"n%2F2\"[2*7 + (n-1)12]\r\n" );
document.write( "\r\n" );
document.write( "S12 = \"12%2F2\"[14 + (12-1)12]\r\n" );
document.write( "\r\n" );
document.write( "S12 = 6[14 + 11×12]\r\n" );
document.write( "\r\n" );
document.write( "S12 = 6[14 + 132]\r\n" );
document.write( "\r\n" );
document.write( "S12 = 6[146]\r\n" );
document.write( "\r\n" );
document.write( "S12 = 876\r\n" );
document.write( "\r\n" );
document.write( "Then we substitute n = 13 to see if that runs over 1000.\r\n" );
document.write( "\r\n" );
document.write( "To prove it we substitute n = 12\r\n" );
document.write( "\r\n" );
document.write( "Sn = \"n%2F2\"[2*7 + (n-1)12]\r\n" );
document.write( "\r\n" );
document.write( "S13 = \"13%2F2\"[14 + (13-1)12]\r\n" );
document.write( "\r\n" );
document.write( "S13 = \"13%2F2\"[14 + 12×12]\r\n" );
document.write( "\r\n" );
document.write( "S13 = \"13%2F2\"[14 + 144]\r\n" );
document.write( "\r\n" );
document.write( "S13 = \"13%2F2\"[158]\r\n" );
document.write( "\r\n" );
document.write( "S13 = 1027\r\n" );
document.write( "\r\n" );
document.write( "This runs over 1000, so we now have proved that\r\n" );
document.write( "\r\n" );
document.write( "the answer is 12.\r\n" );
document.write( "\r\n" );
document.write( "---------------\r\n" );
document.write( "\r\n" );
document.write( "For the second part of the problem, it is done exactly the same\r\n" );
document.write( "way, just use 2000 instead of 1000.\r\n" );
document.write( "\r\n" );
document.write( "The answer is  18.\r\n" );
document.write( "\r\n" );
document.write( "Edwin
\n" ); document.write( "
\n" );