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 $p$, $q$, 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, $d^2 = d \cdot p$.

Since $p$, $q$, and $r$ are unequal, $d \neq 0$.

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 $3d : 5d : 4d = 3 : 5 : 4$.

Answer: $3:5:4$