Four numbers in A.P.: (a-3d,\ a-d,\ a+d,\ a+3d)

Sum: (4a=32\implies a=8)

Sum of squares: [ (8-3d)^2+(8-d)^2+(8+d)^2+(8+3d)^2=276 ] [ 2[(8-3d)^2+(8-d)^2]=276 ] [ (64-48d+9d^2)+(64-16d+d^2)=138 ] [ 128-64d+10d^2=138 ] [ 10d^2-64d-10=0 ] [ 5d^2-32d-5=0 ] [ d=\dfrac{32\pm\sqrt{1024+100}}{10}=\dfrac{32\pm\sqrt{1124}}{10} ] Wait, recheck.

Actually expand properly: [ 2(64-48d+9d^2+64-16d+d^2)=276 ] [ 2(128-64d+10d^2)=276 ] [ 256-128d+20d^2=276 ] [ 20d^2-128d-20=0 ] [ 5d^2-32d-5=0 ] [ d=\dfrac{32\pm\sqrt{1024+100}}{10}=\dfrac{32\pm\sqrt{1124}}{10}=\dfrac{32\pm2\sqrt{281}}{10} ] Hmm that seems messy. Common approach uses (a-3d/2) etc.

Let numbers be (a-3d,a-d,a+d,a+3d): Sum of squares = (4a^2+20d^2=276) With (a=8): (4\cdot64+20d^2=276\implies256+20d^2=276\implies20d^2=20\implies d^2=1\implies d=\pm1)

Numbers: (5,7,9,11)

Answer: (5,7,9,11)