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

1 Answer
Sep 17, 2017

17

Explanation:

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 4th 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).