Question

If l, m, n are in arithmetic progression, then straight line lx + my +n = 0 will pass through point
(a)(-1, 2)
(b)(1, -2)
(c)(1, 2)
(d)(2, 1)

Answer 1

Hint: Convert l, m, n into general arithmetic progression terms,
l, m, n = l, (l + d), (l + 2d)
Find a relation between them to get a relation between l, m, n and then convert the relation to a straight line with algebraic coefficients.

Complete step-by-step answer:
A sequence is said to be in arithmetic progression, if the difference of any two successive numbers is a constant. For example, the sequence 1, 2, 3, 4, ... is an arithmetic progression with common difference 1. 
Now we have l, m, n that are in arithmetic progression, So we have that l is the first term of the sequence.
Let us assume d is a common difference.
So from above we can say the value of m to be:
m = l + d …..(1)
And now we need value of n, it will be:
n = l + 2d …..(2)
By simplifying equation (1),
By subtracting l on both sides, we get:
m – l = l + d – l
By simplifying, we get:
m – l = d
d = m – l …..(3)
Substituting the value of d in equation (3) into the equation (2), we get:
n = l + 2(m - l)
By simplifying, we get:
n = l + 2m - 2l = 2m –l
By adding l on both sides, we get:
n + l = 2m – l + l
By cancelling the common terms, we get:
n + l = 2m
By subtracting 2m on both sides, we get:
n + l – 2m = 2m – 2m = 0
By arranging the above equation, we get:
l – 2m + n = 0 …..(4)
Given equation in question is:
 Line - lx + my +n = 0 …..(5)
By comparing equation (4) and equation (5), we get:
x = 1 and y = -2
So the point (x, y) through which it passes is (1, -2).
\[\therefore \] If l, m, n are in arithmetic progression then the line lx + my + n = 0 will pass through (1, -2).
So option (b) is correct

Note: Alternate method is to convert them into other forms of arithmetic progression.
l, m, n = (m – d), m, (m + d).
This will give your result with less calculation.