Does anyone know how to solve this find four geometric means between 32 and 1?

Question

I need to know the steps

Answer 1

let them be 32, a, b, c, d, 1

then we have a^2 = 32b, b^2 = ac, c^2=bd, d^2 = c

c^2 = bd = b sqr(c) or c^(3/2) = b or c = b ^(2/3)

b^2 = ac means b^2 = a b^(2/3) or b^(4/3) = a

but a^2 = 32b or b^(8/3) = 32 b or b^(5/3)= 32 or b = 32^(3/5) = 8
(since 32^(1/5)=2 as 2^5=32)

b=8 => a= 16 and hence c=4 and d=2 ....

Answer 2

I am not sure the question is worded correctly. If it is so, try this.

A. 1) list 2,3,4,5,6,7,8 in a column. 2) Take the log of each number. 3) Add the logs together. 4) Divide by the number of values in the column (7 for this column and the last one, 8 for the two middle columns). 5) Then take the antilog of the answer. That is the first geometric mean.

B. Do the same process to the remaining three lists: 9 thru 16; 17 thru 24, and 25 thru 31.


This is one way to get 4 geometric means for the values between 1 and 32 (your question said "between", so 1 and 32 have to be excluded).