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.