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
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
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
2, 4, 8, 16,...
If
-1, 1, -1, 1,...
So the sum of the possible
Random Bonus
-
It is one of the few know Fermat primes, being a prime number of the form
2^(2^n)+1 , specifically2^(2^2)+1 . Pierre de Fermat conjectured that all such numbers are prime, but it fails for2^(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 from1 ) which is expressible as the sum of a square and a cube in2 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).