given a,b,c are in gp means b/a = c/b (common ratio)
=> b2 = ac ...............@1
now given ap =bq = cr
p loga = q logb = r logc = k(say) [taking log on every values]
so, a = 10 k/p b= 10 k/q c= 10 k/r
putting these values in @1.
10 2k/q = 10 ( k/p + k/r)
2k/q = k/p + k/r
2/q = 1/p + 1/r (cancelling k from both sides)
1/q -1/p = 1/r -1/q (writing above eqn in other way)
which means that 1/p , 1/q , 1/r is in AP.
now we know that
if p,q,r is in AP than i/p,1/q,1/r is in HP.
so, as 1/p,1/q,1/r is in AP than p,q,r is in HP..
answer is option C.