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If $l,m,n$ are in A.P. then the lines represented by $lx + my + n = 0$ are concurrent at the point:
(1)$\left( {1,2} \right)$
(2) $\left( {2,4} \right)$
(3) $\left( { - 2,1} \right)$
(4) $\left( {1, - 2} \right)$

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Last updated date: 20th Jun 2024
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Answer
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Hint: Use the given data that $l,m,n$ are in Arithmetic progression, which means that the common difference of each consecutive term is constant, which gives an equation. On comparing the obtained equation with the given equation, we get the required point.

Complete step-by-step answer:
It is given in the problem that $l,m,n$are in Arithmetic progression and we have to find the point where the lines represented by $lx + my + n = 0$are concurrent.
As given that $l,m,n$ are in Arithmetic progression, it means that the common difference of each congruent term is constant. That is,
$m - l = n - m$
We will simplify the obtained equation by taking the same term of one side.
$ \Rightarrow m + m = n + l$
$ \Rightarrow 2m = n + l$
Now, we shift all the terms on the left-hand side.
$\Rightarrow l - 2m - n = 0$
Now, we compare the above equation with the given equation:
$\Rightarrow lx + my + n = 0$
On comparing both the equation, the coefficient of $l$ given that:
$\Rightarrow x = 1$
On comparing both the equation, the coefficient of $m$ gives that:
$\Rightarrow y = - 2$
So, we have the values $x = 1$and$y = - 2$. Then the obtained point $\left( {x,y} \right)$ is $\left( {1, - 2} \right)$.
Hence, we can see that the line $lx + my + n = 0$ passes through the point $\left( {1, - 2} \right)$. Therefore, the lines represented by $lx + my + n = 0$ are concurrent at the point $\left( {1, - 2} \right)$.
Hence, option (4) is correct.

Note: The concurrent lines are the lines that intersect at a single point and that point is asking for the question. One line is given to us and one line is formed using the given data, then the point of intersection of these points is the point where the formed lines are concurrent.