If the lines `ax + ky + 10 =0, bx + (k+1) y + 10 =0 and cx + (k +2)y + 10=0` are concurrent, then

Options

A. a,b,c are in GP
B. a,b,c are in HP
C. a,b,c are in AP
D. `(a+b^(2))=c`

Answer

C

Explanation

Since , the given lines are concurrent.
`|{:(a,k,10),(b,k+1,10),(c,k+2,10):}|=0`
` Rightarrow 10|{:(a,k,1),(b,k+1,1),(c,k+2,1):}|=0`
Applying ` R_(2)toR_(2)-R_(1) and R_(3)to R_(3)-R_(1)`
`10|{:(a,k,1),(b-a,1,0),(c-a,2,0):}|=0`
` Rightarrow 10[1(2b-2a-c+a)}=0 Rightarrow 2b=a+c`
Thus, a,b and c are in AP.