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Mathematics problems in the first volume of senior one.
Solution: Because the first three numbers are arithmetic progression, we can assume that these four numbers are X, x+d, x+2d, Y,

Then: (x+2d)? = (x+d)y

x+y=37

x+d + x+2d =36

So: (x+2d)? =(x+d)(37-x); 2x+3d=36。

Therefore: 2x+3d+ 1=37.

So: (x+2d)? = (x+d)( 2x+3d+ 1-x)

Namely: (x+2d)? = (x+d)( x+3d+ 1)

So: x? +4dx+4d? =x? +3dx+x+dx+3d? +d

Therefore: x=d? -d becomes 2x+3d=36.

D: 2d? +d-36=0

Therefore: d=4 or d=-9/2.

(1) when d=4, x=d? -d = 12, so: y=25.

These four numbers are 12, 16, 20, 25.

(2) when d=-9/2, x=d? -d = 99/4, so: y=49/4.

These four numbers are 99/4, 8 1/4, 63/4 and 49/4.