Question

If the \((a+1) t h, 7 t h\) and \((b+1)\) th terms of an \(A P\) are in \(G P\) with \(a, 6, b\) being in \(H P\), then 4 th term of this \(A P\) is (a) \(-\frac{7}{2}\) (b) \(\frac{7}{2}\) (c) 0 (d) 3

Short answer

Answer: The 4th term of the AP is \(\frac{7}{2}\).

Step-by-step solution

Understanding AP, GP, and HP

In an arithmetic progression, the difference between consecutive terms is constant, and this constant is called the common difference. In a geometric progression, the ratio between consecutive terms is constant, and this constant is called the common ratio. In a harmonic progression, the reciprocals of the terms form an AP.

Define expressions for the AP terms

Let the AP have a first term \(A\) and a common difference \(d\). The terms of the AP follow this pattern: \((a+1)\)th term: \(A + ad\) \(7\)th term: \(A + 6d\) \((b+1)\)th term: \(A + bd\)

Set up a GP relationship for the three given terms

Since the \((a+1)\)th, \(7\)th, and \((b+1)\)th terms of the AP are in GP, we can set up an equation by taking the product of the first and third terms and equating it to the square of the second term: \((A + ad)(A + bd) = (A + 6d)^2\)

Set up a HP relationship for a, 6, and b

We have to remember that for numbers in an HP (\(a\), \(6\), \(b\)), their reciprocals are in AP. Denote the reciprocal of \(a\) as \(x\), the reciprocal of \(6\) as \(y\), and the reciprocal of \(b\) as \(z\). Then: \(x = \frac{1}{a}\) \(y = \frac{1}{6}\) \(z = \frac{1}{b}\) Since \(x\), \(y\), and \(z\) are in AP, we have the following relationship: \(2y = x + z\)

Substituting the HP expressions into the GP equation

Substitute the reciprocal expressions for \(a\) and \(b\) into the GP equation: \((A + \frac{d}{x})(A + \frac{d}{z}) = (A + 6d)^2\) Substitute the right-hand side of the HP equation \((2y = x + z)\): \((A + \frac{d}{x})(A + \frac{d}{2y - x}) = (A + 6d)^2\)

Solving the GP equation for x

Now, we will solve the equation for \(x\) using the fact that \(y = \frac{1}{6}\): \((A+\frac{d}{x})(A+\frac{d}{\frac{1}{3}-x})=(A+6d)^2\)

Find the fourth term of AP

Given the information in the problem and found x, we can determine the fourth term of the AP: \(4\)th term = \(A + 3d\) From step 3, we found the \(7\)th term of the AP is \(A + 6d\) Now we know the \(4\)th term is half of the \(7\)th term (since HP has the reciprocals in an AP): \(4\)th term = \(\frac{1}{2}(A + 6d) = \frac{7}{2}\) Therefore, the \(4\)th term of the AP is \(\boxed{\frac{7}{2}}\) (option b).

Key concepts

Geometric Progression (GP)

In a geometric progression (GP), each term is derived from the previous one by multiplying it with a fixed constant known as the common ratio. This characteristic structure means that if you have the first term and the common ratio, you can easily find any term in the GP by repeatedly multiplying by the common ratio.
For example, consider a GP with the first term, denoted as \(a\), and a common ratio \(r\). The sequence of terms follows:

• First term: \(a\)
• Second term: \(ar\)
• Third term: \(ar^2\)
• Fourth term: \(ar^3\)

And so on. In this exercise, the given terms of an arithmetic progression (AP) need to satisfy the condition of being in GP, which means we need to ensure the product of the outer terms equals the square of the middle term. This specific condition creates a bridge between sequences, where the strategies used in analyzing GPs can often simplify complicated terms.

Harmonic Progression (HP)

A harmonic progression (HP) is a sequence of numbers formed by taking the reciprocals of an arithmetic progression (AP). This means that if you have numbers \(a, b, c\) in an HP, their reciprocals \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) will be in AP.
Applying this to our exercise, the numbers \(a, 6, b\) are given in HP. To analyze them, we first take the reciprocals and verify they form an AP. This provides the equation \(2\left(\frac{1}{6}\right) = \frac{1}{a} + \frac{1}{b}\).
Such relations help find missing values in complex sequences when combined with GP conditions, providing a powerful tool for problem-solving. Understanding HPs can make analyzing intricate relationships more manageable, as it converts problems into a simpler arithmetic framework that is often easier to work with.

Common Difference

The common difference is a foundational concept in the study of arithmetic progressions (AP). It refers to the constant difference between consecutive terms in the sequence. This value remains the same throughout the AP and serves as a key to finding any term in the sequence if the first term is known.
When a sequence follows the pattern of an AP, and is defined by the first term \(A\) and a common difference \(d\), the \(n\)th term can be determined by the formula:

• \(n\)th term: \(A + (n-1)d\)

This formula is crucial in resolving the exercise's problem since it allows us to express certain terms mathematically, particularly when setting up relationships with GP or HP. Identifying the common difference facilitates understanding of the sequence's behavior and helps relate it to other mathematical constructs, such as the geometric progression found in the problem. Thus, recognizing and utilizing the common difference is essential in multi-sequence problems, providing clarity and easing computation.