Question 971266
Given, a,b,c are in GP, 
So, let common factor be k such that b=ak and c=ak^2.
Now for checking the H.P. of given terms,
{{{1/log(a,x)-1/log(b,x)}}}
={{{log(x,a)-log(x,b)}}}
={{{log(x,(a/b))}}}
={{{log(x,(a/(a*k)))}}}
={{{log(x,(1/k))}}}  -(i)
Again,
{{{1/log(b,x)-1/log(c,x)}}}
={{{log(x,b)-log(x,c)}}}
={{{log(x,(b/c))}}}
={{{log(x,((a*k)/(a*k^2)))}}}
={{{log(x,(1/k))}}}  -(ii)
So, from (i) and (ii),
{{{1/log(a,x)-1/log(b,x) = 1/log(b,x)-1/log(c,x)}}}
Hence, {{{log(a,x)}}},{{{log(b,x)}}} and {{{log(c,x)}}} are in H.P.