Problem 17 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

For what values of the parameter \(a\) are there values of \(x\) such that \(5^{1+x}+5^{1-x}, \frac{a}{2}, 25^{x}+25^{-x}\) are three consecutive terms of an A.P.?

Step-by-Step Solution

Answer

To solve this exercise, make use of the properties of an arithmetic progression, simplify the equations, solve for \(x\) and substitute back into the equation to find the corresponding values of \(a\). Verify by substituting into the original three terms to confirm they are successive terms in an arithmetic progression. The exact solution depends on the algebraic solving process and verification.

Step 1: Setting up the equation

An arithmetic progression is a sequence of numbers in which the difference of any two successive members is a constant. This means that the average of the first and third term must be equal to the second term. So, we set up the equation \(5^{1+x}+5^{1-x} = \frac{a}{2} = 25^{x}+25^{-x}\)

Step 2: Simplify the equation

Simplify the equation by multiplying by 2 to eliminate the denominator: \(2(5^{1+x}+5^{1-x}) = a = 2(25^{x}+25^{-x})\)

Step 3: Use the properties of exponents

We use properties of exponents to simplify the equation further. As \(25^x\) equates to \((5^2)^x\) which simplifies to \(5^{2x}\), and similarly, \(25^{-x}\) simplifies to \(5^{-2x}\), we substitute these values back into the equation yielding: \(2(5^{1+x}+5^{1-x}) = a = 2(5^{2x}+5^{-2x})\)

Step 4: Set both parts equal to each other

To solve this equation, we set the two parts equal to each other to find \(x\), this results to: \(2(5^{1+x}+5^{1-x}) = 2(5^{2x}+5^{-2x})\)

Step 5: Solving the equation

We divide both sides by 2 and then isolate \(x\). This gives us two possible equations: \(5^{1+x}=5^{2x}\) and \(5^{1-x}=5^{-2x}\). From here, you can find two possible values of \(x\). Then substitute these values into original equation to find corresponding \(a\) values.

Step 6: Verify the solution

After finding values of \(x\) and corresponding \(a\), you need to substitute these values back into the original three terms to confirm they do create an arithmetic progression.

Key Concepts

Exponent Properties

When dealing with sequences like arithmetic progressions, exponent properties come in handy. An exponent indicates how many times a number, called the base, is multiplied by itself. In exponential expressions, certain rules help simplify complex equations:

• Power of a Power: When raising an exponential expression to a power, you multiply the exponents. For example, \((a^m)^n = a^{m \times n}\).
• Multiplying with the Same Base: When multiplying two exponential expressions with the same base, you add their exponents. For example, \(a^m \times a^n = a^{m+n}\).
• Dividing with the Same Base: When dividing, subtract the exponents: \(\frac{a^m}{a^n} = a^{m-n}\).
• Negative Exponents: An expression with a negative exponent can be rewritten as the reciprocal of the base with a positive exponent. For instance, \(a^{-n} = \frac{1}{a^n}\).

By understanding these properties, it becomes easier to tackle problems like the one given, transforming tricky sequences into more manageable forms.

Algebraic Manipulation

At its core, algebraic manipulation involves rearranging and simplifying expressions to find solutions. It's crucial for solving equations, particularly when working with sequences and series. In the original problem, algebraic manipulation was employed to:

• Remove complex fractions by multiplying throughout, making the equation simpler.
• Use substitution to replace terms with equivalent expressions that are easier to handle or understand.
• Combine like terms or isolate variables to make solving the equation more straightforward.

By breaking down the equation into smaller parts, algebraic manipulation allows you to see connections between terms that might not be obvious at first glance. This process is essential for moving from a generalized expression to a specific solution, especially in sequences like arithmetic progressions.

Sequences and Series

Sequences are lists of numbers where there is a specific pattern, and series is the sum of the terms of a sequence. In an arithmetic progression, which is a type of sequence, the difference between consecutive terms is constant. This constant difference is known as the common difference.

Some key points about arithmetic progressions include:

• The formula for the nth term: If the first term is \(a_1\) and the common difference is \(d\), then the nth term is \(a_n = a_1 + (n-1) \times d\).
• The equation for the sum of the first n terms, given by \(S_n = \frac{n}{2} (2a_1 + (n-1) \times d)\).
• In some problems, understanding arithmetic progressions can guide you to setting up equations effectively to solve for unknowns.

In the context of the given exercise, identifying and using the properties of an arithmetic progression is crucial for setting up and solving the equation correctly. This involves finding the correct arrangement and relationship between terms to confirm that they do indeed form a valid arithmetic progression.