How do you find 33 geometric means between 33 and 14881488 ?

1 Answer
Jul 25, 2016

3, color(red)(6root(4)(31)), color(green)(12sqrt(31)), color(blue)(24(root(4)(31))^3), 14883,6431,1231,24(431)3,1488

Explanation:

We are looking for 33 numbers color(red)(a), color(green)(b), color(blue)(c)a,b,c such that:

color(red)(a)a is the geometric mean of 33 and color(green)(b)b

color(green)(b)b is the geometric mean of color(red)(a)a and color(blue)(c)c

color(blue)(c)c is the geometric mean of color(green)(b)b and 14881488

That will make the following sequence into a geometric one:

3, color(red)(a), color(green)(b), color(blue)(c), 14883,a,b,c,1488

If the common ratio is rr then we must have:

1488 = 3 r^41488=3r4

So:

r^4 = 1488/3 = 496 = 2^4*31r4=14883=496=2431

in order that the geometric means be Real and positive, we need to choose the principal 44th root to find:

r = 2root(4)(31)r=2431

Hence color(red)(a), color(green)(b), color(blue)(c)a,b,c are:

3*2root(4)(31) = color(red)(6root(4)(31))32431=6431

6root(4)(31)*2root(4)(31) = color(green)(12sqrt(31))64312431=1231

12sqrt(31)*2root(4)(31) = color(blue)(24(root(4)(31))^3)12312431=24(431)3