Let p, q and r be three unequal numbers such that p, q and r are in AP. If (q – p), (r – q) and p are in GP, then (p + q) : (q + r) : (r + p) equals

Since p, q, r are in AP, we have 2q = p + r, so q − p = r − q = d (common difference). For (q − p), (r − q), p to be in GP: (r − q)² = (q − p)·p, i.e., d² = d·p, giving p = d. Then q = p + d = 2p and r = p + 2d = 3p. So p + q = 3p, q + r = 5p, r + p = 4p. The ratio (p+q):(q+r):(r+p) = 3:5:4. Wait, this gives 3:5:4, but with the order asked (p+q):(q+r):(r+p) = 3p:5p:4p = 3:5:4. However, reviewing carefully: the question asks the ratio in that order, yielding 3:5:4. Re-examining option matching: 3:5:4 corresponds to option (c). Correct answer is option (c).