Analysis of Harmonic Progression (HP), Arithmetic Progression (AP), and Geometric Progression (GP)

The question provides three terms \(\frac{1}{b+c}, \frac{1}{c+a},\frac{1}{a+b}\) that are stated to be in Harmonic Progression (HP). We need to determine the correctness of two statements based on this information: whether \(a, b, c\) are in Arithmetic Progression (AP) and whether \((b + c)^2, (c + a)^2, (a + b)^2\) are in Geometric Progression (GP).

A fundamental property of a Harmonic Progression is that the reciprocals of the terms in HP are in Arithmetic Progression (AP).

Given that \(\frac{1}{b+c}, \frac{1}{c+a},\frac{1}{a+b}\) are in HP, their reciprocals must be in AP. The reciprocals are:

• \(b+c\)
• \(c+a\)
• \(a+b\)

Therefore, \(b+c, c+a, a+b\) are in AP.

Deriving the Condition for a, b, c

For three terms \(X, Y, Z\) to be in AP, the middle term \(Y\) must be the average of the first term \(X\) and the third term \(Z\). Mathematically, this is expressed as \(2Y = X + Z\).

Applying this condition to the terms \(b+c, c+a, a+b\) which are in AP, we have:

\(2(c+a) = (b+c) + (a+b)\)

Expanding both sides:

\(2c + 2a = b + c + a + b\)

Combining like terms:

\(2c + 2a = a + c + 2b\)

Subtract \(a\) and \(c\) from both sides:

\(2a - a + 2c - c = 2b\)

\(a + c = 2b\)

This equation \(a + c = 2b\) is the defining condition for three numbers \(a, b, c\) to be in Arithmetic Progression.

Checking Statement 1: a, b, c are in AP

Statement 1 says that \(a, b, c\) are in AP. Our derivation from the initial HP condition led directly to the condition \(a + c = 2b\), which is the definition of \(a, b, c\) being in AP.

Thus, statement 1 is correct.

Checking Statement 2: (b + c)\({}^2\), (c + a)\({}^2\), (a + b)\({}^2\) are in GP

Statement 2 says that \((b + c)^2, (c + a)^2, (a + b)^2\) are in GP. For three terms \(P, Q, R\) to be in GP, the square of the middle term \(Q\) must be equal to the product of the first term \(P\) and the third term \(R\). Mathematically, this is expressed as \(Q^2 = PR\).

Applying this condition to the terms \((b + c)^2, (c + a)^2, (a + b)^2\), we must check if:

\(( (c+a)^2 )^2 = (b+c)^2 \cdot (a+b)^2\)

Taking the square root of both sides (assuming positive values, or considering the absolute value):

\((c+a)^2 = (b+c)(a+b)\)

From our analysis of the HP condition, we know that \(a+c = 2b\). Substituting this into the potential GP condition:

\((2b)^2 = (b+c)(a+b)\)

\(4b^2 = ab + b^2 + ac + bc\)

Rearranging the terms:

\(4b^2 - b^2 - ab - ac - bc = 0\)

\(3b^2 - ab - ac - bc = 0\)

Since \(a, b, c\) are in AP, we can write \(a = b-d'\) and \(c = b+d'\) for some common difference \(d'\). Substitute these into the equation:

\(3b^2 - (b-d')b - (b-d')(b+d') - b(b+d') = 0\)

\(3b^2 - (b^2 - bd') - (b^2 - (d')^2) - (b^2 + bd') = 0\)

\(3b^2 - b^2 + bd' - b^2 + (d')^2 - b^2 - bd' = 0\)

Combine like terms:

\((3 - 1 - 1 - 1)b^2 + (bd' - bd') + (d')^2 = 0\)

\(0 \cdot b^2 + 0 + (d')^2 = 0\)

\((d')^2 = 0\)

This equation \((d')^2 = 0\) implies \(d' = 0\). If the common difference \(d'\) of the AP (\(a, b, c\)) is 0, it means \(a=b=c\).

So, the condition that \((b + c)^2, (c + a)^2, (a + b)^2\) are in GP is only true if \(a=b=c\). The initial HP condition (\(\frac{1}{b+c}, \frac{1}{c+a},\frac{1}{a+b}\) are in HP) implies that \(a, b, c\) are in AP, but it does not necessarily imply that \(a=b=c\) (unless the common difference of the AP is specifically 0). The statement "are in GP" needs to hold true whenever the initial HP condition holds, not just in the specific case where \(a=b=c\).

Since statement 2 is not always true based solely on the fact that \(\frac{1}{b+c}, \frac{1}{c+a},\frac{1}{a+b}\) are in HP (it requires the additional condition \(a=b=c\)), statement 2 is incorrect in the general case implied by the question.

Conclusion

Based on the analysis, only statement 1 is correct.

Statement 1: \(a, b, c\) are in AP (Correct)

Statement 2: \((b + c)^2, (c + a)^2, (a + b)^2\) are in GP (Incorrect in the general case)

Revision Table: Properties of Progressions

Additional Information: Relationship Between AP, GP, HP

When three numbers \(a, b, c\) are related, they can form different progressions under specific conditions:

• If \(a, b, c\) are in AP, then \(b = \frac{a+c}{2}\). \(b\) is the Arithmetic Mean (AM) of \(a\) and \(c\).
• If \(a, b, c\) are in GP (with positive terms), then \(b = \sqrt{ac}\). \(b\) is the Geometric Mean (GM) of \(a\) and \(c\).
• If \(a, b, c\) are in HP (with positive terms), then \(b = \frac{2ac}{a+c}\). \(b\) is the Harmonic Mean (HM) of \(a\) and \(c\).

For any two positive numbers \(a\) and \(c\), the AM, GM, and HM are related by the inequality \(AM \ge GM \ge HM\). Equality holds if and only if \(a=c\).

In the problem, we found that if \(\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}\) are in HP, then \(b+c, c+a, a+b\) are in AP. This directly shows that \(c+a\) is the Arithmetic Mean of \(b+c\) and \(a+b\).

The condition \(a+c=2b\) derived is the specific relationship between \(a, b, c\) that arises from the given HP condition. This condition means that \(a, b, c\) form an AP.

The second statement about \((b+c)^2, (c+a)^2, (a+b)^2\) being in GP implies \((c+a)^2 = (b+c)(a+b)\). We showed this requires the common difference of the AP (\(a, b, c\)) to be zero, meaning \(a=b=c\). This is a stricter condition than \(a, b, c\) being in AP.