<PRE><B>Question 260234</B>
Let  n = 5
     d = 1
    40 = the sum of 5 consecutive integers
Sn = n/2(2a&lt;sub&gt;1&lt;/sub&gt;+(n-1)d)
40 = 5/2(2a&lt;sub&gt;1&lt;/sub&gt;+(5-1)1)
40 = 5/2(2a&lt;sub&gt;1&lt;/sub&gt;+4)
40 = (5/2) 2(a&lt;sub&gt;1&lt;/sub&gt;+2)
40 = 5(a&lt;sub&gt;1&lt;/sub&gt;+2)
8  = a&lt;sub&gt;1&lt;/sub&gt;+2
6  = a&lt;sub&gt;1&lt;/sub&gt;

The smallest number is 6.
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