aruthmetic progressions

a,b,c are in A.P.

So,

2b=a+c

So,

 b=(a+c)/2

So, the equation of the line ax+by+c=0

can be written as

ax+(a+c/2)y+c=0

So we get,

2ax+(a+c)y+2c=0

 

Now if (1,-2) has to lie on this line, we must have that the coordinates of this point  always satisfy the equation .

So, putting  x=1, y= -2 in the equation of the line we get,

2a-2(a+c)+2c

=2(a+c)-2(a+c)

=0

So we see that the point (1,-2) satisfies the equation of the line whatever may be the values of  a,b,c as long as they are in A.P.

Hence proved.