If M is the mean of x1,x2,x3,x4,x5 and x6,prove that (x1−M)+(x2−M)+(x3−M)+(x4−M)+(x5−M)+(x6−M)=0
∴ M is the mean of x1,x2,x3,x4,x5 and x6 Then M=x1+x2+x3+x4+x5+x66⇒x1+x2+x3+x4+x5+x6=6M ...(i)Now L.H.S.=(x1−M)+(x2−M)+(x3−M)+(x4−M)+(x5−M)+(x6−M)=x1+x2+x3+x4+x5+x6−M−M−M−M−M−M=x1+x2+x3+x4+x5+x6−6M=6M−6M [From (i)]=0=R.H.S.