Question 1132426
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Let x be the 5th term of the sequence; let a be the common difference.<br>
(NOTE: Choosing x to be the 5th term of the sequence puts it halfway between the 3rd and 7th terms; this will simplify the algebra required to solve the problem.)<br>
Then the 2nd term is x-3a; the 9th term is x+4a; the 3rd term is x-2a, and the 7th term is x+2a.  The given information is then<br>
{{{(x-3a)+(x+4a) = 25}}}  -->  {{{2x+a = 25}}}
{{{(x-2a)+(x+2a) = 20}}}  -->  {{{2x = 20}}}<br>
From those two equations, it is easy to solve for x and a, enabling us to write the first 5 terms of the sequence.