If x=112+132+152+...., y=112+322+132+342+.. and z=112−122+132−142+...., then
y−x=[112+322+132+342+...]−[112+132+152+...]
=3[122+142+162+...]......(1)
Similarly,
x−z=[112+132+152+....]−[112−122+132−142+....]
=122+142+162........(2)
From (1) and (2),
y−x=3(x−z)⟹y+3z=4x
⇒(y6)+(z2)=2(x3) (dividing both sides by 6)
Therefore y6,x3andz2 are in A.P