Question

If \( a+1, 2a+1, 4a-1 \) are in arithmetic progression, then the value of \( a \) is:

• A. 1
• B. 2
• C. 3
• D. 4
• E. 5

Hint:

The "Arithmetic Mean" property (\( 2 \times \text{Middle} = \text{Sum of Ends} \)) is the most reliable way to solve for variables in an A.P. It avoids the extra step of calculating the common difference separately.

Answer 1

The Correct Option is B

Concept: For any three terms \( x, y, z \) to be in arithmetic progression, the difference between consecutive terms must be equal (\( y - x = z - y \)). Alternatively, this is equivalent to the property that twice the middle term equals the sum of the outer terms: \( 2y = x + z \).

Step 1: Setting up the A.P. property equation.
Given the terms \( x = a+1 \), \( y = 2a+1 \), and \( z = 4a-1 \). Applying the property \( 2y = x + z \): \[ 2(2a + 1) = (a + 1) + (4a - 1) \]

Step 2: Solving for \( a \).
Expand and simplify both sides of the equation: \[ 4a + 2 = 5a + 0 \] Rearrange to isolate \( a \): \[ 5a - 4a = 2 \] \[ a = 2 \]