



 
Hi, I think the problem says
If the common difference is $d$ then here are a few of the terms
The sum of these terms is 1030 so \[4 + (4+d) + (4+2d) + \cdot \cdot \cdot + (4 + 19 d) = 1030.\] which simplifies to \[20 \times 4 + d(1 + 2 + \cdot \cdot \cdot + 19) = 1030.\] Solve for $d.$ Penny  


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