Understanding the Problem: AP and HP Series

The problem describes two sequences involving terms a and b. The first sequence a, x, y, z, b is an Arithmetic Progression (AP), and the second sequence a, p, q, r, b is a Harmonic Progression (HP). We are given conditions on the sums of some terms in each sequence and asked to find the value of the product ab.

Analyzing the Arithmetic Progression (AP)

An Arithmetic Progression (AP) is a sequence where the difference between consecutive terms is constant. In the given AP: a, x, y, z, b.

We are given that x + y + z = 15.

Since x, y, and z are consecutive terms in an AP, the middle term y is the arithmetic mean of x and z. That is, \( y = \frac{x+z}{2} \), which implies \( x+z = 2y \).

Substituting this into the given sum:

\( (x+z) + y = 15 \)

\( 2y + y = 15 \)

\( 3y = 15 \)

Dividing by 3, we find the value of y:

\( y = 5 \)

Now consider the full sequence a, x, y, z, b which is an AP of 5 terms. In an AP with an odd number of terms, the middle term is the arithmetic mean of the first and the last term. Here, y is the third term, which is the middle term.

So, \( y = \frac{a+b}{2} \).

Substituting the value of y = 5:

\( 5 = \frac{a+b}{2} \)

Multiplying by 2:

\( a+b = 10 \)

We have found a relationship between a and b from the AP condition: a + b = 10.

Analyzing the Harmonic Progression (HP)

A Harmonic Progression (HP) is a sequence where the reciprocals of the terms are in an Arithmetic Progression (AP). In the given HP: a, p, q, r, b.

The reciprocals $\frac{1}{a}, \frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{b}$ are in an AP.

We are given that $\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = \frac{5}{3}$.

Since $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}$ are consecutive terms in an AP, the middle term $\frac{1}{q}$ is the arithmetic mean of $\frac{1}{p}$ and $\frac{1}{r}$. That is, \( \frac{1}{q} = \frac{\frac{1}{p} + \frac{1}{r}}{2} \), which implies \( \frac{1}{p} + \frac{1}{r} = \frac{2}{q} \).

Substituting this into the given sum:

\( (\frac{1}{p} + \frac{1}{r}) + \frac{1}{q} = \frac{5}{3} \)

\( \frac{2}{q} + \frac{1}{q} = \frac{5}{3} \)

\( \frac{3}{q} = \frac{5}{3} \)

Cross-multiplying:

\( 3 \times 3 = 5 \times q \)

\( 9 = 5q \)

Dividing by 5:

\( q = \frac{9}{5} \)

So, $\frac{1}{q} = \frac{5}{9}$.

Now consider the full sequence of reciprocals $\frac{1}{a}, \frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{b}$, which is an AP of 5 terms. Similar to the first AP, the middle term $\frac{1}{q}$ is the arithmetic mean of the first and the last term.

So, \( \frac{1}{q} = \frac{\frac{1}{a} + \frac{1}{b}}{2} \).

Substituting the value of $\frac{1}{q} = \frac{5}{9}$:

\( \frac{5}{9} = \frac{\frac{1}{a} + \frac{1}{b}}{2} \)

Multiplying by 2:

\( \frac{1}{a} + \frac{1}{b} = \frac{10}{9} \)

Combine the terms on the left side by finding a common denominator:

\( \frac{b+a}{ab} = \frac{10}{9} \)

Or equivalently:

\( \frac{a+b}{ab} = \frac{10}{9} \)

Calculating the Value of ab

We have two key relationships derived from the AP and HP conditions:

• From the AP: \( a+b = 10 \)
• From the HP: \( \frac{a+b}{ab} = \frac{10}{9} \)

Now, substitute the value of \( a+b \) from the first equation into the second equation:

\( \frac{10}{ab} = \frac{10}{9} \)

To solve for ab, we can cross-multiply or simply observe that if the numerators are equal (both are 10) and the fractions are equal, then the denominators must also be equal.

Thus,

\( ab = 9 \)

Final Answer Calculation

Based on the analysis of the given Arithmetic Progression and Harmonic Progression, and using the provided sum conditions, we found that \(a+b=10\) and $\frac{a+b}{ab} = \frac{10}{9}$. Substituting the first result into the second allowed us to directly solve for the product ab.

The value of ab is 9.

Revision Table: AP and HP Properties

Additional Information: Means (AM, GM, HM)

The concepts of Arithmetic Progression and Harmonic Progression are related to different types of means. For two positive numbers A and B:

• Arithmetic Mean (AM): \( \text{AM} = \frac{A+B}{2} \). If A, x, B are in AP, then x is the AM of A and B ($x = \frac{A+B}{2}$). In our problem, y is the AM of a and b, and $\frac{1}{q}$ is the AM of $\frac{1}{a}$ and $\frac{1}{b}$.
• Geometric Mean (GM): \( \text{GM} = \sqrt{AB} \). If A, g, B are in GP (Geometric Progression), then g is the GM of A and B ($g^2 = AB$).
• Harmonic Mean (HM): \( \text{HM} = \frac{2}{\frac{1}{A} + \frac{1}{B}} = \frac{2AB}{A+B} \). If A, h, B are in HP, then h is the HM of A and B ($h = \frac{2}{\frac{1}{A} + \frac{1}{B}}$). In our problem, q is the HM of a and b ($q = \frac{2}{\frac{1}{a} + \frac{1}{b}}$). We found $\frac{1}{q} = \frac{\frac{1}{a} + \frac{1}{b}}{2}$, which is equivalent to $q = \frac{2}{\frac{1}{a} + \frac{1}{b}}$.

There is a relationship between these means: For positive numbers, \( \text{AM} \ge \text{GM} \ge \text{HM} \). Also, \( \text{GM}^2 = \text{AM} \times \text{HM} \).

In this problem, we used the properties of terms within AP and HP sequences, specifically the property of the middle term in a sequence with an odd number of terms, rather than directly using the definitions of AM/HM between 'a' and 'b'. However, note that y is the AM of a and b, and q is the HM of a and b.

• $y = \frac{a+b}{2} = \frac{10}{2} = 5$. This matches our finding for y.
• $q = \frac{2}{\frac{1}{a} + \frac{1}{b}} = \frac{2}{\frac{10}{9}} = 2 \times \frac{9}{10} = \frac{18}{10} = \frac{9}{5}$. This matches our finding for q.

Knowing that y is the AM and q is the HM of a and b could provide an alternative path using the $\text{GM}^2 = \text{AM} \times \text{HM}$ relationship if 'a' and 'b' were positive. The geometric mean of a and b would be $\sqrt{ab}$. So, $(\sqrt{ab})^2 = y \times q$. $ab = 5 \times \frac{9}{5} = 9$. This confirms the result and shows the connection to AM, GM, and HM.