Wait — the original assumption about ordering may be wrong. Try symmetric AP: let the four terms be $ a - 3d, a - d, a + d, a + 3d $ — symmetric around $ a $. This is a valid arithmetic progression with common difference $ 2d $, but we can scale. Define as AP with common difference $ 2d' $, but to match convention, let common difference be $ 2d $, so terms: $ a - 3d, a - d, a + d, a + 3d $. Then first: $ a - 3d $, last: $ a + 3d $, sum of squares: $ (a-3d)^2 + (a+3d)^2 = 2a^2 + 18d^2 $. Sum of all: $ 4a $, square: $ 16a^2 $. Set equal: - Abu Waleed Tea