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

Step-by-step solution

Since p , q , and r are in AP, let the common difference be d . Then q - p = d and r - q = d . It is given that (q - p) , (r - q) , and p are in GP. Substituting the values, we get d , d , and p are in GP. Therefore, d^2 = d p . Since p , q , and r are unequal, d 0 . Thus, p = d . Now, q = p + d = 2d and r = p + 2d = 3d . We need to find the ratio (p+q) : (q+r) : (r+p) . Substituting the values of p , q , and r in terms of d : p + q = d + 2d = 3d q + r = 2d + 3d = 5d r + p = 3d + d = 4d The required ratio is 3d : 5