Question 823018
<pre>
10,17,24,...

Let's find out if this arithmetic sequence can have two terms
that differ by 100..  The formula for the nth term of an
arithmetic sequence is

a<sub>n</sub> = a<sub>1</sub> + (n-1)d

The common difference d = 17-10 = 24-17 = 7 and the first term
a<sub>1</sub> = 10.  Substituting:

a<sub>n</sub> = 10 + (n-1)(7)
a<sub>n</sub> = 10 + 7(n-1)
a<sub>n</sub> = 10 + 7n - 7
a<sub>n</sub> = 3 + 7n

Let's find out whether the pth and qth terms 
can differ by 100 for some p and q, where p > q and p and q 
are positive integers.

a<sub>p</sub> = 3 + 7p
a<sub>q</sub> = 3 + 7q

Set their difference &#8799; 100.  (The question mark over the =
indicates that we are not sure that they can be equal.)

 (3 + 7p) - (3 + 7q) &#8799; 100

     3 + 7p - 3 - 7q &#8799; 100

             7p - 7q &#8799; 100

            7(p - q) &#8799; 100

               p - q &#8799; {{{100/7}}}

{{{100/7}}} is not an integer.  However p and q
are integers, and the difference of two integers 
is always an integer.  So p - q cannot equal {{{100/7}}}

So the answer to the question asked is no.
    
Edwin</pre>