The problem asks us to find the ratio $(p+q) : (q+r) : (r+p)$ given that $p, q, r$ are unequal numbers in Arithmetic Progression (AP) and $(q-p), (r-q), p$ are in Geometric Progression (GP).

Understanding AP and GP Conditions

• AP Condition: If $p, q, r$ are in AP, the common difference is constant. Let the common difference be $d$. Then, $q - p = r - q = d$.
  • From this, we get $q = p + d$ and $r = q + d = (p + d) + d = p + 2d$.
• GP Condition: If $(q-p), (r-q), p$ are in GP, the ratio between consecutive terms is constant. This means $(r-q)^2 = (q-p) \times p$.

Solving for p, q, and r

1. Substitute AP relations into GP condition: We know $q-p = d$ and $r-q = d$. Substituting these into the GP condition $(r-q)^2 = (q-p) \times p$, we get: $ d^2 = d \times p $
2. Find the relationship between d and p: Since $p, q, r$ are unequal, the common difference $d$ cannot be zero ($d \neq 0$). Therefore, we can divide the equation $d^2 = dp$ by $d$: $ d = p $
3. Express q and r in terms of p: Now substitute $d=p$ back into the AP relations:
   • $q = p + d = p + p = 2p$
   • $r = p + 2d = p + 2p = 3p$
   So, the numbers are $p, 2p, 3p$. These satisfy the condition of being unequal (assuming $p \neq 0$) and in AP with a common difference $d=p$. They also satisfy the GP condition: $(q-p) = p$, $(r-q) = p$, and $p$. So, $p, p, p$ are indeed in GP since $p^2 = p \times p$.

Calculating the Required Ratio

We need to find the ratio $(p+q) : (q+r) : (r+p)$.

• Calculate each term using $q=2p$ and $r=3p$:
  • $p+q = p + 2p = 3p$
  • $q+r = 2p + 3p = 5p$
  • $r+p = 3p + p = 4p$
• Form the ratio: $ (p+q) : (q+r) : (r+p) = 3p : 5p : 4p $
• Simplify the ratio by dividing by $p$ (since $p \neq 0$): $ 3 : 5 : 4 $

The ratio $(p+q) : (q+r) : (r+p)$ is $3 : 5 : 4$.