I solved this using two hashmaps/dictionaries. The idea is to iterate using the middle element (let's call it i) of the triplet, and maintaining two hashmaps/dictionaries, one which holds all the elements before i, and the other holds all the elements after i. At the beginning of each iteration, remove i from the after hashmap, and at the end of each iteration, add i to the before hashmap.

Here is the solution implemented in python:

def countTriplets(arr, r):
    after = {}
    before = {}
    for a in arr:
        after[a] = after[a] + 1 if a in after else 1
        before[a] = 0
        
    pairs = 0

    for num in arr:
        after[num] -= 1
        af = num * r
        bf = num / r

        if af in after and bf in before:
            pairs += (after[af] * before[bf])
        
            
        before[num] += 1

    return pairs