If the system of linear equations `x+2ay +az =0,x+3by+bz =0 and x+4cy + cz =0` has a non-zero solution, then `a, b, c`

Options

A. are in AP
B. are in GP
C. are in HP
D. satisfy `a+2b+3c=0`

Answer

C

Explanation

Since then system of linear equations has a non zero solution then
`|(1,2a,a),(1,3b,b),(1,4c,c)|=0`
Applyign `R_(2)toR_(2)-R_(1)` and `R_(3)toR_(3)-R_(1)` we get
`=|(1,2a,a),(0,3b-2a,b-a),(0,4c-2a,c-a)|=0`
`implies(3b-2a)(c-a)-(4c-2a)(b-a)=0`
`implies3bc-3ba-2ac+2a^(2)=4bc-2ab-4ac+2a^(2)`
`implies2ac=bc+ab`
On dividing by abc both sides, we get
`2/b=1/a+1/c`
`:.a,b` and `c` are in HP.