`x-3, x-1, 3x-7` form a geometric sequence. What is the sum of all possible fourth terms of such a sequence?

Given geometric sequence:

`x-3, x-1, 3x-7`

The square of the middle term must be equal to the product of the first and last terms.

So:

`x^2-2x+1 = (x-1)^2 = (x-3)(3x-7) = 3x^2-16x+21`

Subtracting `x^2-2x+1` from both ends, this becomes:

`0 = 2x^2-14x+20`

`color(white)(0) = 2(x^2-7x+10)`

`color(white)(0) = 2(x-5)(x-2)`

So:

`x = 5" "` or `" "x = 2`

If `x=5` then the sequence is:

`2, 4, 8, 16,...`

If `x=2` then the sequence is:

`-1, 1, -1, 1,...`

So the sum of the possible `4`th terms is `16+1=17`

Random Bonus

`17` is actually my favourite number, having several interesting properties:

• It is one of the few know Fermat primes, being a prime number of the form `2^(2^n)+1`, specifically `2^(2^2)+1`. Pierre de Fermat conjectured that all such numbers are prime, but it fails for `2^(2^5)+1 = 4294967297 = 641 * 6700417` and no larger Fermat number is known to be prime.

• Because it's a Fermat prime, a regular `17`-sided polygon is one of the few known prime number sided regular polygons constructible with ruler and compasses.

• `17` is the smallest number (apart from `1`) which is expressible as the sum of a square and a cube in `2` distinct ways: `17 = 4^2+1^3 = 3^2+2^3`

• `17` is the number of distinct possible symmetries of wallpaper patterns (i.e. biperiodic tesselations of the plane).