If the system of linear equations ax + ky + 2az = 0 ; bx + ky + 3bz = 0 ; cx + ky + 4cz = 0where k,a,b,c ∈ Rhas non- zero solution then

System has non – zero solution ⇒Δ=0 ⇒ ak2abk3bck4c=0⇒ka12ab13bc14c=0 ⇒a12ab13bc14c=0 Expanding along column C2, we get        -14bc-3bc+14ac-2ac-13ab-2ab=0 ⇒2ac=bc+ab Dividing throughout by abc, we get        2b=1a+1c ⇒a,b,c are in H.P.