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

Sep 17, 2017

$17$

#### Explanation:

Given geometric sequence:

$x - 3 , x - 1 , 3 x - 7$

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

So:

${x}^{2} - 2 x + 1 = {\left(x - 1\right)}^{2} = \left(x - 3\right) \left(3 x - 7\right) = 3 {x}^{2} - 16 x + 21$

Subtracting ${x}^{2} - 2 x + 1$ from both ends, this becomes:

$0 = 2 {x}^{2} - 14 x + 20$

$\textcolor{w h i t e}{0} = 2 \left({x}^{2} - 7 x + 10\right)$

$\textcolor{w h i t e}{0} = 2 \left(x - 5\right) \left(x - 2\right)$

So:

$x = 5 \text{ }$ or $\text{ } x = 2$

If $x = 5$ then the sequence is:

$2 , 4 , 8 , 16 , \ldots$

If $x = 2$ then the sequence is:

$- 1 , 1 , - 1 , 1 , \ldots$

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 ${2}^{{2}^{5}} + 1 = 4294967297 = 641 \cdot 6700417$ 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).