Question

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

• A. $a,b,c$ are in G.P.
• B. $a,b,c$ are in H.P.
• C. $a,b,c$ are in A.P.
• D. ${(a+1)}^2=c$

Answer 1

The correct option is C

$a,b,c$ are in A.P.

Explanation for correct option

Given lines, $ax+ky+10=0,bx+(k+1)y+10=0$ and $cx+(k+2)y+10=0$ are concurrent.

So, the determinant will be zero.

$\Rightarrow \left|\begin{array}{ccc}a& k& 10\\ b& k+1& 10\\ c& k+2& 10\end{array}\right|=0$

Applying the row transformation operations,

$$\begin{gathered} R_2\to R_2-R_1 \\ {R}_3\to R_3-R_1 \end{gathered}$$

$\Rightarrow \left|\begin{array}{ccc}a& k& 10\\ b-a& 1& 0\\ c-a& 2& 0\end{array}\right|=0$

Upon solving, we get,

$$\begin{gathered} \Rightarrow a(0-0)-k(0-0)+10(2b-2a-c+a)=0 \\ \Rightarrow 2b-a-c=0 \\ \Rightarrow a+c=2b \end{gathered}$$

So, the condition obtained is of Arithmetic progression.

Therefore, option (C) is the correct answer.