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a1+a2+a3=12
a1+9;a2+9;a3+9 ciąg geometryczny
a1+a1+r+a1+2r=12
3a1+3r=12
a1+r=4
a1=4-r
4-r+9;4-r+r+9;4-r+2r+9
13-r;13;13+r
13²=(13-r)(13+r)
169=169-r²
r²=0
r=0
musi to więc byc ciąg stały a1=4 r=0