if the system of linear equations
`{:(x+2ay+az=0),(x+3by+bz=0),(x+4cy+cz=0):}`
has a non zero solution then a,b,c are in

Options

A. H.P.
B. G.P.
C. A.P.
D. none of these

Answer

A

Explanation

The given system of linear equations has a non trivial solution. Therefore,
`|(1,4a,a),(1,3b,b),(1,2c,c)| = 0`
`rArr |(1,4a,a),(0,2c - 4a,b -a),(0,2c - 4a,c -a)| = " " [("Applying " R_(2) rarr R_(2) - R_(1)),(" " and R_(3) rarr R_(3) - R_(1))]`
`rArr (3b - 4a) (c -a) - (2c -4a) (b -a) = 0`
`rArr bc + ab - 2ac = 0`
`rArr (1)/(a) + (1)/(c) = (2)/(b) rArr a, b, c` are in H.P.