Question 1196632
<pre>
{{{ a[1] = 1/8 }}}  which is  {{{ (1/8)*k^0}}}  where k is the common ratio

You are asked to insert {{{a[2]}}} thru {{{ a[5] }}} while being given
{{{a[6] = 128 }}}  

{{{ a[6] = (1/8)*k^5 = 128 }}}  -->  {{{ k^5 = 8*128 = 1024 = 2^10 }}}
which implies  {{{ k = 2^2 = 4 }}}

{{{ a[1] = 1/8 }}}
{{{ a[2] = 4(1/8) = 1/2 }}}
{{{ a[3] = 4(1/2) = 2 }}}
{{{ a[4] = 4(2) = 8 }}}
{{{ a[5] = 4(8) = 32 }}}
and as a check
{{{ a[6] = 4(32) = 128 }}}

Hope this helps! 

EDIT: Note to student: I assumed you wanted four elements of a geometric progression, inserting "4 geometric means" does not make sense to me.