How do you find #3# geometric means between #3# and #1488# ?

1 Answer
George C.
Jul 25, 2016

#3, color(red)(6root(4)(31)), color(green)(12sqrt(31)), color(blue)(24(root(4)(31))^3), 1488#

Explanation:

We are looking for #3# numbers #color(red)(a), color(green)(b), color(blue)(c)# such that:

#color(red)(a)# is the geometric mean of #3# and #color(green)(b)#

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

#color(blue)(c)# is the geometric mean of #color(green)(b)# and #1488#

That will make the following sequence into a geometric one:

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

If the common ratio is #r# then we must have:

#1488 = 3 r^4#

So:

#r^4 = 1488/3 = 496 = 2^4*31#

in order that the geometric means be Real and positive, we need to choose the principal #4#th root to find:

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

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

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

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

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

Related questions
See all questions in Geometric Sequences
Impact of this question
2400 views around the world
You can reuse this answer
Creative Commons License