Problem 112 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(p, q, r\) are in AP, then show that \(p t h, q t h\) and \(r t h\) terms of any GP are in GP.

Step-by-Step Solution

Answer

To show that the \(p^{th}\), \(q^{th}\), and \(r^{th}\) terms of a GP are in a GP, we follow these steps: 1. Understand that the given \(p, q, r\) are in an AP, i.e. \(q - p = r - q\). 2. Find \(p^{th}\), \(q^{th}\), and \(r^{th}\) terms of GP: \(a_p = a(r^{p-1})\), \(a_q = a(r^{q-1})\), \(a_r = a(r^{r-1})\). 3. Check if the obtained terms are in GP: \(\frac{a_q}{a_p} = \frac{a_r}{a_q}\). 4. Simplify the equation by cancelling out the terms and using the common base r: \(r^{(q-1)-(p-1)} = r^{(r-1)-(q-1)}\). 5. Use the given Arithmetic Progression condition: \((q-1)-(p-1) = (r-1)-(q-1)\). 6. Compare the simplified equation from Step 4 with the equation obtained in Step 5 and conclude that the \(p^{th}\), \(q^{th}\), and \(r^{th}\) terms of a GP are in a GP.

Step 1: Understand the given information

We are given that p, q, and r are in AP, which means that: \(q - p = r - q\) Now, let a GP be given by the sequence {a, ar, ar², ar³, ...}, where a is the first term, and r is the common ratio. We need to show that the pth, qth, and rth terms of the GP are in GP.

Step 2: Find the pth, qth, and rth terms of GP

Now we will determine the pth, qth, and rth terms of the given GP: \(a_p = a(r^{p-1})\) \(a_q = a(r^{q-1})\) \(a_r = a(r^{r-1})\) We have the general terms for the pth, qth, and rth terms of a GP.

Step 3: Check if the obtained terms are in GP

To prove that the obtained terms are in GP, we need to show that: \(\frac{a_q}{a_p} = \frac{a_r}{a_q}\) With the obtained terms: \(\frac{a(r^{q-1})}{a(r^{p-1})} = \frac{a(r^{r-1})}{a(r^{q-1})}\)

Step 4: Simplify the equation

We will simplify the equation by cancelling out the terms on both sides: \(\frac{r^{q-1}}{r^{p-1}} = \frac{r^{r-1}}{r^{q-1}}\) Dividing both sides by their common base r, we get: \(r^{(q-1)-(p-1)} = r^{(r-1)-(q-1)}\)

Step 5: Use the given Arithmetic Progression condition

As we know that p, q, and r are in AP, we have: \(q - p = r - q\) Now, we can rewrite the above equation as: \((q-1)-(p-1) = (r-1)-(q-1)\)

Step 6: Prove that pth, qth, and rth terms of the GP are in GP

On comparing the simplified equation from Step 4 with the equation obtained in Step 5, we have: \(r^{(q-1)-(p-1)} = r^{(r-1)-(q-1)}\) This implies that the pth, qth, and rth terms of the given GP are in GP, as the common ratio is the same for all three terms. Thus, we have shown that if p, q, and r are in AP, then the pth, qth, and rth terms of any GP are in GP.

Key Concepts

Geometric Progression

Understanding a Geometric Progression (GP) is crucial when tackling problems involving sequences. A GP is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number known as the common ratio. This results in a sequence that grows (or shrinks) exponentially rather than linearly.

Some key characteristics of a GP include:

• Each term is a constant multiple of the previous term.
• The ratio between successive terms is consistent throughout the sequence.
• The formula for the nth term in a GP is given by \(a_n = a \cdot r^{n-1}\), where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the term number.

A GP can be either increasing (when \(r > 1\)), decreasing (when \(0 < r < 1\)), or constant (when \(r = 1\)). In mathematical problems involving GPs, recognizing and applying these characteristics is key to solving problems efficiently.

Common Ratio

The common ratio in a Geometric Progression is the multiplier that transforms each term into the next. It's a fundamental component that defines the GP's behavior and can directly affect whether a sequence converges or diverges.

To find the common ratio \(r\), you simply divide any term by its preceding term in the sequence:
\(r = \frac{a_{n+1}}{a_n}\).

The value of \(r\):

• If \(|r| > 1\), the terms of the sequence will grow larger with each step.
• If \(0 < |r| < 1\), the terms will shrink, approaching zero.
• If \(r = 1\), every term is the same, resulting in a constant sequence.

Understanding the common ratio's impact on a sequence’s nature helps in analyzing and solving progression-related problems, as seen in the original exercise, where the common ratio verifies that terms remain consistent.

Sequence Terms

In sequences, whether in Arithmetic or Geometric Progressions, each element is referred to as a term. These terms are essentially the building blocks of the sequence, organized in a specific order.

In a GP, each term can be represented using the formula \(a_n = a \cdot r^{n-1}\). This formula reveals how the initial term \(a\) and the common ratio \(r\) contribute to the entire sequence:

• The first term, \(a\), sets the stage for the entire progression, influencing all subsequent terms.
• Each term retains the identity of the sequence by maintaining the consistent ratio \(r\) with the term before it.

Analyzing the position and the value of sequence terms plays a pivotal role in understanding and solving sequence-related exercises.

In our exercise context, showing that specific terms like \(p^{th}, q^{th},\) and \(r^{th}\) continue to form a GP proves that the consistent application of the common ratio keeps the sequence intact following the outlined properties of the progression.