Edexcel IAL January 2022 P4 (WMA14/01) Q8 Proof by Contradiction
Three consecutive terms in a sequence of real numbers are given by
\( k, 1 + 2k \ \text{and}\ 3 + 3k \)
where k is a constant.
Use proof by contradiction to show that this sequence is not a geometric sequence.
(5)
Solution to Question: Proof by Contradiction
Key Concepts Used:
- Proof by Contradiction: Assume the sequence is geometric.
- Definition of Geometric Sequence: The ratio of consecutive terms, \( r \), is constant, so \( \frac{u_2}{u_1} = \frac{u_3}{u_2} \).
- Quadratic Discriminant: Use of \( b^2 – 4ac \) to show that the resulting quadratic equation has no real solutions for \( k \).
Step-by-Step Solution:
- Form the Assumption
Assume the sequence \( k, 1 + 2k, 3 + 3k \) is a geometric sequence.
- Set Ratios Equal
By the definition of a geometric sequence, the common ratio must be equal:
\( r = \frac{1 + 2k}{k} = \frac{3 + 3k}{1 + 2k} \) - Solve the Equation for \( k \)
Cross-multiply and expand:
\( k(3 + 3k) = (1 + 2k)(1 + 2k) \)
\( 3k + 3k^2 = 1 + 4k + 4k^2 \) - Form a Quadratic Equation
Rearrange the equation to set it equal to zero:
\( 0 = (4k^2 – 3k^2) + (4k – 3k) + 1 \)
\( k^2 + k + 1 = 0 \) - Check for Real Solutions (Discriminant)
Calculate the discriminant, \( \Delta = b^2 – 4ac \), where \( a = 1, b = 1, c = 1 \):
\( \Delta = (1)^2 – 4(1)(1) = 1 – 4 = -3 \) - Identify the Contradiction
Since the discriminant \( \Delta = -3 \) is less than zero (\( \Delta < 0 \)), the quadratic equation \( k^2 + k + 1 = 0 \) has no real solutions for \( k \). Since \( k \) must be a real constant for the terms to exist as real numbers, the sequence cannot be geometric. This contradicts the initial assumption.
Final Answer:
Since the assumption that the sequence is geometric leads to a quadratic equation with no real solutions for k, the assumption is incorrect. Therefore, the sequence is not a geometric sequence.