Given: ax + by + c = 0 is a variable straight line, where a, b, and c are the 1st, 3rd, and 7th terms of an increasing arithmetic progression (A.P.).

To prove: The variable straight line always passes through a fixed point that lies on the equation x^2 + y^2 = 13.

Proof:

1. Definition of an Arithmetic Progression (A.P.):
An arithmetic progression is a sequence of numbers in which the difference between consecutive terms is constant. Let the first term be a, the common difference be d, and the nth term be Tn. Then the arithmetic progression can be represented as: a, a + d, a + 2d, ..., a + (n-1)d.

2. Representation of the straight line:
The equation of a straight line in the form ax + by + c = 0 can be rewritten as y = (-a/b)x - (c/b). Here, the slope of the line is -a/b and the y-intercept is -c/b.

3. Relationship between terms of A.P. and the straight line:
The given straight line has the coefficients a, b, and c, which are the 1st, 3rd, and 7th terms of an increasing arithmetic progression. Therefore, we can represent these terms as a, a + 2d, and a + 6d.

4. Relationship between slope and y-intercept:
The slope of the line is given by -a/b. Substituting the values of a and b, we have:
Slope = -(a/b) = -(a/(a+2d)) = -1/(1+2d/a)

The y-intercept of the line is given by -c/b. Substituting the value of c, we have:
Y-intercept = -(c/b) = -(a + 6d)/(a + 2d)

5. Finding the fixed point:
To find the fixed point, we need to find the intersection of the line with the equation x^2 + y^2 = 13. Substituting y = (-a/b)x - (c/b) in the equation, we get:
x^2 + [(-a/b)x - (c/b)]^2 = 13

Simplifying the equation, we obtain:
(1 + (a^2/b^2))x^2 + 2ac/b^2 x + (c^2/b^2 - 13) = 0

This is a quadratic equation in x. For the line to intersect the circle, the discriminant of the quadratic equation should be zero.
Discriminant = 4a^2c^2/b^4 - 4(1 + (a^2/b^2))(c^2/b^2 - 13) = 0

Simplifying further, we have:
4a^2c^2 - 4b^2(c^2 - 13 - a^2) = 0
4a^2c^2 - 4b^2c^2 + 4b^2a^2 - 4b^2(13 - a^2) = 0
4a^2c^2 -