This question involves finding the common difference of an Arithmetic Progression (AP) given certain terms and the condition that the product of two specific terms is minimized. Let's break down the problem step-by-step.

An Arithmetic Progression is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference, denoted by \(d\).

Let the first term of the AP be \(a_1\), the third term be \(a_3\), and the fifth term be \(a_5\).

According to the question:

• The first term is \(p\), so \(a_1 = p\).
• The third term is \(q\), so \(a_3 = q\).
• The fifth term is \(3\), so \(a_5 = 3\).
• The common difference is \(d\).

The general form of the n-th term of an AP is given by \(a_n = a_1 + (n-1)d\).

Using this formula, we can write the given terms in terms of the first term (\(p\)) and the common difference (\(d\)):

• \(a_1 = p\)
• \(a_3 = a_1 + (3-1)d = p + 2d\). Since \(a_3 = q\), we have \(q = p + 2d\).
• \(a_5 = a_1 + (5-1)d = p + 4d\). Since \(a_5 = 3\), we have \(p + 4d = 3\).

We are given that the product \(pq\) is minimum. To minimize \(pq\), we first need to express \(p\) and \(q\) in terms of the common difference \(d\).

From the equation for the fifth term, \(p + 4d = 3\), we can express \(p\) in terms of \(d\):

\(p = 3 - 4d\)

Now substitute this expression for \(p\) into the equation for \(q\):

\(q = p + 2d = (3 - 4d) + 2d = 3 - 2d\)

Now we can form the product \(pq\):

\(pq = (3 - 4d)(3 - 2d)\)

Let's expand this expression:

\(pq = 3 \times 3 + 3 \times (-2d) + (-4d) \times 3 + (-4d) \times (-2d)\)

\(pq = 9 - 6d - 12d + 8d^2\)

\(pq = 8d^2 - 18d + 9\)

The product \(pq\) is a quadratic function of \(d\) in the form \(Ad^2 + Bd + C\), where \(A=8\), \(B=-18\), and \(C=9\). This is a parabola that opens upwards because the coefficient of \(d^2\) (which is \(A=8\)) is positive. The minimum value of such a quadratic function occurs at the vertex.

The d-coordinate of the vertex of a parabola \(Ad^2 + Bd + C\) is given by the formula \(d = \frac{-B}{2A}\).

Using this formula, we can find the value of \(d\) that minimizes \(pq\):

\(d = \frac{-(-18)}{2 \times 8}\)

\(d = \frac{18}{16}\)

Simplifying the fraction, we get:

\(d = \frac{9}{8}\)

Thus, the value of the common difference \(d\) that minimizes the product \(pq\) is \(\frac{9}{8}\).

Arithmetic Progression Terms and Product Minimization

Here's a summary of how we related the terms and found the expression for the product:

Product \(pq = (3 - 4d)(3 - 2d) = 8d^2 - 18d + 9\).

Finding the Minimum Value of the Product

The product \(pq\) is a quadratic expression \(f(d) = 8d^2 - 18d + 9\). The graph of this function is a parabola opening upwards. The minimum value occurs at the vertex.

The formula for the axis of symmetry (which gives the d-coordinate of the vertex) for a quadratic \(Ad^2 + Bd + C\) is \(d = \frac{-B}{2A}\).

For \(f(d) = 8d^2 - 18d + 9\), we have \(A=8\) and \(B=-18\).

Value of \(d\) for minimum \(pq\):

\(d = \frac{-(-18)}{2(8)} = \frac{18}{16} = \frac{9}{8}\).

Final Result for Common Difference

The value of the common difference \(d\) for which the product \(pq\) is minimum is \(\frac{9}{8}\).

Revision Table: AP Concepts

Additional Information: Minimizing Quadratic Functions

A quadratic function is a polynomial of degree 2, generally written as \(f(x) = Ax^2 + Bx + C\), where A, B, and C are constants and \(A \neq 0\).

• If \(A > 0\), the parabola opens upwards, and the function has a minimum value at its vertex.
• If \(A < 0\), the parabola opens downwards, and the function has a maximum value at its vertex.

The x-coordinate of the vertex is given by \(x = \frac{-B}{2A}\). This value of x gives the minimum (if \(A>0\)) or maximum (if \(A<0\)) value of the function.

In our problem, the product \(pq\) was a quadratic function of \(d\), specifically \(8d^2 - 18d + 9\). Since the coefficient of \(d^2\) is \(8\) (which is greater than 0), the function has a minimum value. The value of \(d\) where this minimum occurs is given by \(d = \frac{-(-18)}{2(8)} = \frac{18}{16} = \frac{9}{8}\).