Problem 116 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

The \(r\) th, \(s\) th and \(t\) th terms of a certain G.P. are \(R, S\) and \(T\) respectively. Prove that \(R^{s-t} \cdot S^{t-r} \cdot T^{r-s}=1\).

Step-by-Step Solution

Answer

Given that R, S, and T are the rth, sth, and tth terms of a G.P., we can represent them as: \(R = a\cdot r^{r-1}\), \(S = a\cdot r^{s-1}\), and \(T = a\cdot r^{t-1}\) Substituting these expressions into the given relationship and simplifying, we get: \(a^{s-t} \cdot a^{t-r} \cdot a^{r-s} \cdot r^{r(s-t)-1(s-t)+s(t-r)-1(t-r)+t(r-s)-1(r-s)} = 1\) This simplifies to: \(a^{0} \cdot r^{0} = 1\) Thus, proving that \(R^{s-t} \cdot S^{t-r} \cdot T^{r-s} = 1\).

Step 1: Express R, S, and T in terms of the common ratio

Since R, S, and T are the rth, sth, and tth terms of the G.P., we can express them in terms of the initial term 'a' and the common ratio 'r': \(R = a\cdot r^{r-1}\) \(S = a\cdot r^{s-1}\) \(T = a\cdot r^{t-1}\)

Step 2: Substitute the terms expressions in the given relationship

Substitute the expressions for R, S, and T we found in the previous step into the given relationship: \(R^{s-t} \cdot S^{t-r} \cdot T^{r-s} = 1\) \((a\cdot r^{r-1})^{s-t} \cdot (a\cdot r^{s-1})^{t-r} \cdot (a\cdot r^{t-1})^{r-s} = 1\)

Step 3: Simplify and solve

Now we simplify the equation and solve for the relationship: \(a^{s-t} \cdot r^{r(s-t)-1(s-t)} \cdot a^{t-r} \cdot r^{s(t-r)-1(t-r)} \cdot a^{r-s} \cdot r^{t(r-s)-1(r-s)} = 1\) Combine the terms with the same base: \(a^{s-t} \cdot a^{t-r} \cdot a^{r-s} \cdot r^{r(s-t)-1(s-t)+s(t-r)-1(t-r)+t(r-s)-1(r-s)}=1 \) Merge the exponents of the terms: \(a^{0} \cdot r^{0} = 1\) Evaluate the exponents: 1 = 1 Since the given relationship results in a true statement, the proof stands verified: \(R^{s-t} \cdot S^{t-r} \cdot T^{r-s} = 1\)

Key Concepts

Exponential Equations

Exponential equations are mathematical expressions where variables appear in exponents. They are critical when dealing with sequences like geometric progressions (GP). In a GP, each term after the first is obtained by multiplying the previous one by a fixed, non-zero number called the common ratio. This relationship allows us to express terms using exponents.

Consider a geometric sequence with initial term \(a\) and common ratio \(r\). The \(n\)th term is found by the formula:

• \(T_n = a \cdot r^{n-1}\)

This formula illustrates that the term's position \(n\) is part of the exponent. When solving exponential equations, it's crucial to manipulate them to have the same base wherever possible. This simplifies calculation, as seen in the solution where each term for \(R\), \(S\), and \(T\) is expressed in terms of \(a\) and \(r\).

By substituting these expressions into the equation \(R^{s-t} \cdot S^{t-r} \cdot T^{r-s} = 1\), we use properties like \(a^0 = 1\) and rules for combining exponents (e.g., \(x^a \cdot x^b = x^{a+b}\)) to solve the problem. The key was aligning terms to reflect the same base, leading to elimination of variable dependencies and a proof that \(1 = 1\).

Proof Techniques

Proofs are logical arguments that confirm the truth of a mathematical statement. The solution to the given exercise uses a particular proof technique, where mathematical expressions undergo transformation and simplification to demonstrate equality.

In this process, identifying shared components, such as the common ratio and initial term in the GP, is a vital starting point. The proof employs:

• Substitution: Replace variables with expressions derived from known relationships in the GP.
• Simplification: Use algebraic identities and rules to reduce complexity, such as combining like terms and manipulating exponents.
• Verification: Show that after all transformations, the equation holds true, delivering the meaningful conclusion that the initial assertion \(R^{s-t} \cdot S^{t-r} \cdot T^{r-s} = 1\) is accurate.

Proofs like these are incredibly effective in not just solving problems but also showcasing a thorough understanding of mathematical concepts and relationships, reinforcing the assurance that calculations align with theoretical principles.

Mathematical Induction

Mathematical induction is a powerful method of proving that a statement is true for all natural numbers. Although not directly used in solving this specific exercise, understanding induction can enrich comprehension of proofs related to sequences and progressions.

Induction typically involves:

• Base Case: Prove the statement for the initial value, often \(n = 1\).
• Inductive Step: Assume the statement is true for some arbitrary \(n = k\), then prove it for \(n = k+1\).

While the problem at hand was tackled through algebraic manipulation, knowing induction aids in grasping more formal and generalized proofs in sequence-based scenarios. For example, proving the general formula for the \(n\)th term of a GP might employ induction to establish that all terms are reachable and valid.

This method is particularly valued in mathematics for its logical depth and capacity to bridge individual instances and general cases, enhancing our ability to definitively prove assertions about infinite sets or sequences.