#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 #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).