<PRE><B>Question 730737</B>
Call the first term x and the second term y.&lt;P&gt;
1: x&lt;P&gt;
2: y&lt;P&gt;
3: x+y&lt;P&gt;
4: x+y + y = x+2y (added the previous two terms).&lt;P&gt;
5: 2x+3y&lt;P&gt;
6: 3x+5y&lt;P&gt;
7: 5x+8y&lt;P&gt;
8: 8x + 13y = 390
9: 13x+21y&lt;P&gt;
The problem states that the second term, y, is greater than the first term, x.  They are both positive and both integers.&lt;P&gt;
Try a few:  y = x+1 gives term 8 as:&lt;P&gt;
8x + 13(x+1) = 390&lt;P&gt;
8x + 13x + 13 = 390&lt;P&gt;
21x = 377&lt;P&gt;
x = 377/21 which is not an integer.  So y cannot be x+1.&lt;P&gt;
Try y = x+2:&lt;P&gt;
8x + 13(x+2) = 390&lt;P&gt;
8x +13x +26 = 390&lt;P&gt;
21x = 364&lt;P&gt;
364/21 = not an integer.&lt;P&gt;
Now there&#39;s a pattern.  In each case (y+n), x= (390-13n)/21.  It must be an integer.&lt;P&gt;
n = 3, 4, 5, 6, 7, 8 don&#39;t work.  But n=9 works.&lt;P&gt;
8x + 13(x+9)= 390&lt;P&gt;
8x+13x + 117 = 390&lt;P&gt;
21x = 273 &lt;P&gt;
x = 273/21 = (390-13*9)/21 = 13.&lt;P&gt;
x = 13 and y = 13+9 = 22&lt;P&gt;
The 9th term is 13x + 21y = 13(13)+21(22) = 169 + 462 = 631
&lt;P&gt;Hope the solution helped.  Sometimes you need more than a solution.  Contact fcabanski@hotmail.com for online, private tutoring, or personalized problem solving (quick for groups of problems.)</PRE>