Question 1044745
If {{{3}}} geometric means are inserted between {{{120}}} and {{{15/2}}} , those five numbers form a geometric sequence.
Also, all five terms are positive, meaning that so is the common ratio.
If {{{r>0}} is the common ratio, the five terms of that sequence are
{{{b[1]=120}}} , {{{b[2]=120*r}}} , {{{b[3]=120*r^2}}} , {{{b[4]=120*r^3}}} 
, and {{{b[5]=120*r^4=15/2}}} .
So,
{{{120*r^4=15/2}}}<-->{{{r^4=(15/2)(1/120)}}}<-->{{{r^4=1/16}}}<-->{{{r=1/2}}}
The thirds of those geometric means can be calculated as
{{{b[4]=120*(1/2)^3}}}--->{{{b[4]=120/8}}}--->{{{b[4]=highlight(15)}}} ,
or more easily as {{{b[4]=b[5](1/r)=(15/2)*2=15}}} .
 
Alternatively, we can calculate the geometric mean of {{{120}}} and {{{15/2}}} , 
which would be the second of the three means to be inserted, as
{{{sqrt(120*(15/2))=sqrt(60*15)=sqrt(15*4*15)=15*2=30}}} .
Then we can calculate the third of the three means to be inserted as the geometric mean of {{{30}}} and {{{15/2}}} ;
{{{sqrt(30*(15/2))=sqrt(15*15)=15}}} .