Analyzing Harmonic and Geometric Progressions

The question asks about the relationship between the terms \(a, b, c\) if the sequence \(\frac{a+b}{2}, b, \frac{b+c}{2}\) is in Harmonic Progression (HP).

Let's first understand what a Harmonic Progression (HP) is. A sequence of non-zero numbers is said to be in HP if the reciprocals of the terms are in Arithmetic Progression (AP).

Given that \(\frac{a+b}{2}, b, \frac{b+c}{2}\) are in HP, their reciprocals must be in AP. The reciprocals are \(\frac{2}{a+b}, \frac{1}{b}, \frac{2}{b+c}\).

If three terms \(P, Q, R\) are in AP, then the middle term is the average of the other two, i.e., \(Q = \frac{P+R}{2}\) or \(2Q = P+R\).

Applying the AP condition to the reciprocals \(\frac{2}{a+b}, \frac{1}{b}, \frac{2}{b+c}\), we get:

\[ 2 \times \frac{1}{b} = \frac{2}{a+b} + \frac{2}{b+c} \] \[ \frac{2}{b} = 2 \left( \frac{1}{a+b} + \frac{1}{b+c} \right) \]

Divide both sides by 2 (assuming \(b \neq 0\), which must be true for the term \(b\) to be in HP):

\[ \frac{1}{b} = \frac{1}{a+b} + \frac{1}{b+c} \]

Now, let's combine the terms on the right side by finding a common denominator:

\[ \frac{1}{b} = \frac{(b+c) + (a+b)}{(a+b)(b+c)} \] \[ \frac{1}{b} = \frac{a+2b+c}{(a+b)(b+c)} \]

Cross-multiply:

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

Expand both sides of the equation:

Left side: \( (a+b)(b+c) = a(b+c) + b(b+c) = ab + ac + b^2 + bc \) Right side: \( b(a+2b+c) = ab + 2b^2 + bc \)

So, the equation becomes:

\[ ab + ac + b^2 + bc = ab + 2b^2 + bc \]

Subtract \(ab\) and \(bc\) from both sides of the equation:

\[ ac + b^2 = 2b^2 \]

Subtract \(b^2\) from both sides:

\[ ac = b^2 \]

This condition, \(b^2 = ac\), is the defining property of a Geometric Progression (GP). A sequence of non-zero numbers \(a, b, c\) is in GP if the ratio of consecutive terms is constant, i.e., \(\frac{b}{a} = \frac{c}{b}\), which simplifies to \(b^2 = ac\).

Therefore, if \(\frac{a+b}{2}, b, \frac{b+c}{2}\) are in HP, then \(a, b, c\) are in GP.

Let's look at the given options:

• Option 1: a, b, c are in AP (means \(2b = a+c\)) - This is not what we found.
• Option 2: a, b, c are in GP (means \(b^2 = ac\)) - This matches our result.
• Option 3: a + b, b + c, c + a are in GP - This is not directly derived from our result for a, b, c.
• Option 4: a + b, b + c, c + a are in AP - This is not directly derived from our result for a, b, c.

Based on our derivation, the correct statement is that \(a, b, c\) are in GP.

Revision Table: Sequence Progressions

Additional Information: Properties of Progressions

Understanding the properties of Arithmetic Progression (AP), Geometric Progression (GP), and Harmonic Progression (HP) is crucial for solving problems involving sequences. Here are a few key points:

• In an AP, terms increase or decrease by a constant value (common difference, \(d\)). The \(n\)-th term is \(a_n = a_1 + (n-1)d\).
• In a GP, terms are multiplied by a constant value (common ratio, \(r\)). The \(n\)-th term is \(a_n = a_1 r^{n-1}\).
• HP does not have a simple formula for the sum of terms or the \(n\)-th term like AP or GP. Problems are usually solved by converting them to AP problems using reciprocals.
• The harmonic mean of two numbers \(a\) and \(c\) is \(H = \frac{2}{\frac{1}{a}+\frac{1}{c}} = \frac{2ac}{a+c}\). If \(a, b, c\) are in HP, \(b\) is the harmonic mean of \(a\) and \(c\).
• The geometric mean of two positive numbers \(a\) and \(c\) is \(G = \sqrt{ac}\). If \(a, b, c\) are in GP (and are positive), \(b\) is the geometric mean of \(a\) and \(c\).
• The arithmetic mean of two numbers \(a\) and \(c\) is \(A = \frac{a+c}{2}\). If \(a, b, c\) are in AP, \(b\) is the arithmetic mean of \(a\) and \(c\).
• For positive numbers \(a\) and \(c\), there is a relationship between the means: \(A \ge G \ge H\). Also, \(G^2 = AH\).