Understanding the Problem

The problem states that four numbers, p, q, r, and s, are in Geometric Progression (G.P.). We need to determine the type of progression for three related terms: \(\frac{1}{{{p^2} + {q^2}}}\), \(\frac{1}{{{q^2} + {r^2}}}\), and \(\frac{1}{{{r^2} + {s^2}}}\).

A Geometric Progression (G.P.) is a sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

Properties of Geometric Progression (G.P.)

If p, q, r, and s are in G.P., let the common ratio be \(k\). Then we can express the terms in relation to the first term p:

• \(q = pk\)
• \(r = qk = (pk)k = pk^2\)
• \(s = rk = (pk^2)k = pk^3\)

So, the terms are \(p, pk, pk^2, pk^3\).

Calculating the Given Terms

Now let's substitute these relationships into the three given terms:

The first term is \(\frac{1}{{{p^2} + {q^2}}}\).

Substitute \(q = pk\):

\[ \frac{1}{{{p^2} + {{(pk)}^2}}} = \frac{1}{{{p^2} + {p^2k^2}}} = \frac{1}{{{p^2}(1 + k^2)}} \]

The second term is \(\frac{1}{{{q^2} + {r^2}}}\).

Substitute \(q = pk\) and \(r = pk^2\):

\[ \frac{1}{{{{(pk)}^2} + {{(pk^2)}^2}}} = \frac{1}{{{p^2k^2} + {p^2k^4}}} = \frac{1}{{{p^2k^2}(1 + k^2)}} \]

The third term is \(\frac{1}{{{r^2} + {s^2}}}\).

Substitute \(r = pk^2\) and \(s = pk^3\):

\[ \frac{1}{{{{(pk^2)}^2} + {{(pk^3)}^2}}} = \frac{1}{{{p^2k^4} + {p^2k^6}}} = \frac{1}{{{p^2k^4}(1 + k^2)}} \]

The three terms are \(\frac{1}{{{p^2}(1 + k^2)}}\), \(\frac{1}{{{p^2k^2}(1 + k^2)}}\), and \(\frac{1}{{{p^2k^4}(1 + k^2)}}\).

Checking for Geometric Progression (G.P.)

Three terms \(a, b, c\) are in G.P. if the ratio of consecutive terms is constant, i.e., \(\frac{b}{a} = \frac{c}{b}\). Let's check the ratios for our terms.

Ratio of the second term to the first term:

\[ \frac{\frac{1}{{{p^2k^2}(1 + k^2)}}}{\frac{1}{{{p^2}(1 + k^2)}}} = \frac{1}{{{p^2k^2}(1 + k^2)}} \times {p^2}(1 + k^2) = \frac{{p^2}(1 + k^2)}{{p^2k^2}(1 + k^2)} \]

Assuming \(p \neq 0\) and \(1 + k^2 \neq 0\) (which is always true for real k), we can cancel the common factors:

\[ \text{Ratio 1} = \frac{1}{k^2} \]

Ratio of the third term to the second term:

\[ \frac{\frac{1}{{{p^2k^4}(1 + k^2)}}}{\frac{1}{{{p^2k^2}(1 + k^2)}}} = \frac{1}{{{p^2k^4}(1 + k^2)}} \times {p^2k^2}(1 + k^2) = \frac{{p^2k^2}(1 + k^2)}{{p^2k^4}(1 + k^2)} \]

Assuming \(p \neq 0\), \(k \neq 0\), and \(1 + k^2 \neq 0\), we can cancel the common factors:

\[ \text{Ratio 2} = \frac{1}{k^2} \]

Since Ratio 1 is equal to Ratio 2, the three terms form a Geometric Progression (G.P.) with a common ratio of \(\frac{1}{k^2}\).

Comparison with A.P. and H.P.

For terms to be in Arithmetic Progression (A.P.), the difference between consecutive terms must be constant. This is not the case here as the ratios are constant.

For terms to be in Harmonic Progression (H.P.), their reciprocals must be in A.P. Let's look at the reciprocals of the three terms:

• Reciprocal 1: \(p^2(1 + k^2)\)
• Reciprocal 2: \(p^2k^2(1 + k^2)\)
• Reciprocal 3: \(p^2k^4(1 + k^2)\)

Let's check if these reciprocals are in A.P. The difference between the second and first reciprocal is:

\[ p^2k^2(1 + k^2) - p^2(1 + k^2) = p^2(1 + k^2)(k^2 - 1) \]

The difference between the third and second reciprocal is:

\[ p^2k^4(1 + k^2) - p^2k^2(1 + k^2) = p^2k^2(1 + k^2)(k^2 - 1) \]

These differences are only equal if \(k^2 - 1 = 0\) (i.e., \(k = \pm 1\)) or if \(p^2(1 + k^2) = 0\) (which means \(p=0\) or \(k^2=-1\), neither of which is typical for a non-trivial G.P. sequence). In general, for any \(k \neq \pm 1\), the differences are not equal. Therefore, the reciprocals are not in A.P., which means the original terms are not in H.P.

Conclusion

Based on the calculations, the ratio between consecutive terms \(\frac{1}{{{p^2} + {q^2}}}\), \(\frac{1}{{{q^2} + {r^2}}}\), and \(\frac{1}{{{r^2} + {s^2}}}\) is constant (\(\frac{1}{k^2}\)). Thus, these terms are in Geometric Progression (G.P.).