The problem provides information about two sequences, one in Arithmetic Progression (AP) and another in Harmonic Progression (HP).

Analyzing the Arithmetic Progression (AP)

We are given that a, x, y, z, b are in AP. Let the common difference be $d$. The terms can be written as:

• First term = a
• Second term = x = a + d
• Third term = y = a + 2d
• Fourth term = z = a + 3d
• Fifth term = b = a + 4d

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

Substituting the terms of the AP:

$(a+d) + (a+2d) + (a+3d) = 15$

$3a + 6d = 15$

Dividing by 3, we get: $a + 2d = 5$.

Notice that $a + 2d$ is the third term, y. So, we have found that $\boxed{y = 5}$.

Alternatively, in an AP of 5 terms (an odd number of terms), the middle term (3rd term) is the average of the first and last terms. Thus, $y = \frac{a+b}{2}$.

Since y = 5, we have $5 = \frac{a+b}{2}$, which implies $\boxed{a+b = 10}$.

Analyzing the Harmonic Progression (HP)

We are given that a, p, q, r, b are in HP. By definition, if terms are in HP, their reciprocals are in AP. So, $\frac{1}{a}, \frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{b}$ are in AP.

Let the common difference of this new AP be $D$. The terms in this AP are:

• First term = $\frac{1}{a}$
• Second term = $\frac{1}{p} = \frac{1}{a} + D$
• Third term = $\frac{1}{q} = \frac{1}{a} + 2D$
• Fourth term = $\frac{1}{r} = \frac{1}{a} + 3D$
• Fifth term = $\frac{1}{b} = \frac{1}{a} + 4D$

We are given that $p^{-1} + q^{-1} + r^{-1} = \frac{5}{3}$. This is the same as $\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = \frac{5}{3}$.

Substituting the terms of the reciprocal AP:

$(\frac{1}{a} + D) + (\frac{1}{a} + 2D) + (\frac{1}{a} + 3D) = \frac{5}{3}$

$\frac{3}{a} + 6D = \frac{5}{3}$.

Similar to the first AP, in the reciprocal AP of 5 terms, the middle term ($\frac{1}{q}$) is the average of the first ($\frac{1}{a}$) and last ($\frac{1}{b}$) terms:

$\frac{1}{q} = \frac{\frac{1}{a} + \frac{1}{b}}{2}$.

Also, $\frac{1}{q}$ is the arithmetic mean of $\frac{1}{p}$ and $\frac{1}{r}$, so $\frac{2}{q} = \frac{1}{p} + \frac{1}{r}$.

Using the given sum: $\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = \frac{5}{3}$.

Substitute $\frac{2}{q}$ for $\frac{1}{p} + \frac{1}{r}$: $\frac{2}{q} + \frac{1}{q} = \frac{5}{3}$.

$\frac{3}{q} = \frac{5}{3}$.

This gives $\boxed{q = \frac{9}{5}}$. Therefore, $\frac{1}{q} = \frac{5}{9}$.

Using the middle term property $\frac{1}{q} = \frac{\frac{1}{a} + \frac{1}{b}}{2}$, we have $\frac{5}{9} = \frac{\frac{1}{a} + \frac{1}{b}}{2}$.

Multiplying by 2 gives $\frac{1}{a} + \frac{1}{b} = \frac{10}{9}$.

We can rewrite $\frac{1}{a} + \frac{1}{b}$ as $\frac{a+b}{ab}$. So, $\frac{a+b}{ab} = \frac{10}{9}$.

Solving for 'a' and 'b'

From the AP analysis, we found $a+b = 10$.

Substitute $a+b=10$ into the equation $\frac{a+b}{ab} = \frac{10}{9}$:

$\frac{10}{ab} = \frac{10}{9}$.

This implies $ab = 9$.

Now we have a system of two equations with two variables:

• $a+b = 10$
• $ab = 9$

Consider a quadratic equation whose roots are a and b. This equation is $t^2 - (a+b)t + ab = 0$.

Substituting the values, we get $t^2 - 10t + 9 = 0$.

Factoring the quadratic equation: $(t-1)(t-9) = 0$.

The roots are $t=1$ and $t=9$. Therefore, the possible values for a and b are {1, 9}.

Finding the Common Difference of the Reciprocal AP

The reciprocal AP is $\frac{1}{a}, \frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{b}$. We know $\frac{1}{q} = \frac{5}{9}$.

Let's consider the two cases for {a, b}:

Case 1: a = 1, b = 9

The reciprocal AP starts with $\frac{1}{a} = \frac{1}{1} = 1$ and ends with $\frac{1}{b} = \frac{1}{9}$.

The sequence is $1, \frac{1}{p}, \frac{5}{9}, \frac{1}{r}, \frac{1}{9}$. This is an AP.

The first term is $A_1 = 1$, the fifth term is $A_5 = \frac{1}{9}$.

In an AP, $A_n = A_1 + (n-1)D$. For the 5th term: $A_5 = A_1 + (5-1)D = A_1 + 4D$.

$\frac{1}{9} = 1 + 4D$.

$4D = \frac{1}{9} - 1 = \frac{1-9}{9} = -\frac{8}{9}$.

$D = -\frac{8}{9} \times \frac{1}{4} = -\frac{2}{9}$.

Now we can find $\frac{1}{p}$ and $\frac{1}{r}$ using this common difference $D = -\frac{2}{9}$ and $A_1 = \frac{1}{a} = 1$:

• $\frac{1}{p} = A_2 = A_1 + D = 1 + (-\frac{2}{9}) = 1 - \frac{2}{9} = \frac{9-2}{9} = \frac{7}{9}$. So, $p = \frac{9}{7}$.
• $\frac{1}{r} = A_4 = A_1 + 3D = 1 + 3(-\frac{2}{9}) = 1 - \frac{6}{9} = 1 - \frac{2}{3} = \frac{3-2}{3} = \frac{1}{3}$. So, $r = 3$.

We already found $q = \frac{9}{5}$.

Let's verify the sum condition: $\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = \frac{7}{9} + \frac{5}{9} + \frac{1}{3} = \frac{7}{9} + \frac{5}{9} + \frac{3}{9} = \frac{7+5+3}{9} = \frac{15}{9} = \frac{5}{3}$. This matches the given information.

Now calculate pqr:

$pqr = \frac{9}{7} \times \frac{9}{5} \times 3 = \frac{9 \times 9 \times 3}{7 \times 5} = \frac{81 \times 3}{35} = \frac{243}{35}$.

Case 2: a = 9, b = 1

The reciprocal AP starts with $\frac{1}{a} = \frac{1}{9}$ and ends with $\frac{1}{b} = \frac{1}{1} = 1$.

The sequence is $\frac{1}{9}, \frac{1}{p}, \frac{5}{9}, \frac{1}{r}, 1$. This is an AP.

The first term is $A_1 = \frac{1}{9}$, the fifth term is $A_5 = 1$.

$A_5 = A_1 + 4D$.

$1 = \frac{1}{9} + 4D$.

$4D = 1 - \frac{1}{9} = \frac{9-1}{9} = \frac{8}{9}$.

$D = \frac{8}{9} \times \frac{1}{4} = \frac{2}{9}$.

Now find $\frac{1}{p}$ and $\frac{1}{r}$ using this common difference $D = \frac{2}{9}$ and $A_1 = \frac{1}{a} = \frac{1}{9}$:

• $\frac{1}{p} = A_2 = A_1 + D = \frac{1}{9} + \frac{2}{9} = \frac{3}{9} = \frac{1}{3}$. So, $p = 3$.
• $\frac{1}{r} = A_4 = A_1 + 3D = \frac{1}{9} + 3(\frac{2}{9}) = \frac{1}{9} + \frac{6}{9} = \frac{7}{9}$. So, $r = \frac{9}{7}$.

We already found $q = \frac{9}{5}$.

Let's verify the sum condition: $\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = \frac{1}{3} + \frac{5}{9} + \frac{7}{9} = \frac{3}{9} + \frac{5}{9} + \frac{7}{9} = \frac{3+5+7}{9} = \frac{15}{9} = \frac{5}{3}$. This matches the given information.

Now calculate pqr:

$pqr = 3 \times \frac{9}{5} \times \frac{9}{7} = \frac{3 \times 9 \times 9}{5 \times 7} = \frac{27 \times 9}{35} = \frac{243}{35}$.

Conclusion

In both possible cases for the values of a and b (which are 1 and 9), the value of pqr is $\frac{243}{35}$. Therefore, the value of pqr is uniquely determined.

The value of pqr is $\frac{243}{35}$.

Revision Table: AP and HP Properties

Additional Information: Means

For two numbers 'a' and 'b':

• Arithmetic Mean (AM) = $\frac{a+b}{2}$
• Geometric Mean (GM) = $\sqrt{ab}$
• Harmonic Mean (HM) = $\frac{2}{\frac{1}{a}+\frac{1}{b}} = \frac{2ab}{a+b}$

If a, y, b are in AP, then y is the AM of a and b ($y = \frac{a+b}{2}$).

If a, q, b are in HP, then q is the HM of a and b ($q = \frac{2ab}{a+b}$).

In our problem:

• y is the middle term of a, x, y, z, b in AP. The 5 terms are equidistant from the end points a and b. While y is the AM of x and z, it's also the AM of a and b since it's the center term of the sequence a, x, y, z, b assuming they are symmetrically placed around the middle term, which they are in an AP. Specifically, y is the AM of a and b because $y = a+2d$ and $b = a+4d$, so $\frac{a+b}{2} = \frac{a + (a+4d)}{2} = \frac{2a+4d}{2} = a+2d = y$. This confirms $y = \frac{a+b}{2}$.
• Similarly, for the HP sequence a, p, q, r, b, their reciprocals $\frac{1}{a}, \frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{b}$ are in AP. The term $\frac{1}{q}$ is the middle term of this reciprocal AP. Therefore, $\frac{1}{q}$ is the AM of $\frac{1}{a}$ and $\frac{1}{b}$. $\frac{1}{q} = \frac{\frac{1}{a} + \frac{1}{b}}{2}$. This is consistent with our derivation $\frac{1}{q} = \frac{5}{9}$ and $\frac{1}{a} + \frac{1}{b} = \frac{10}{9}$. The term q itself is the HM of a and b, i.e., $q = \frac{2ab}{a+b}$. We found $a+b=10$ and $ab=9$, so $q = \frac{2(9)}{10} = \frac{18}{10} = \frac{9}{5}$, which matches our earlier result for q.