Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>Two possible arithmetic progressions with first term \(\dfrac{1}{2}\) are: \(-\dfrac{1}{2}, \dfrac{1}{2}, \dfrac{3}{2}, \ldots\) and \(-\dfrac{3}{2}, \dfrac{1}{2}, \dfrac{5}{2}, \ldots\). Which of the following options are correct?</p>
<p>A) \(-\dfrac{1}{2}, \dfrac{1}{2}, \dfrac{3}{2}, \ldots\) is an A.P.</p>
<p>B) \(-\dfrac{3}{2}, \dfrac{1}{2}, \dfrac{5}{2}, \ldots\) is an A.P.</p>
<p>C) Both are A.P.s with different common differences</p>
<p>D) Neither is an A.P.</p>
Step-by-Step Solution
Key Concept: For an AP with first term a = 1/2, the common difference d can be found using the second term relationship: if second term is b, then d = b - a. Verify that consecutive terms maintain constant difference and identify which sequences validly form APs with this first term.
<p><strong>Step 1: Identify the actual structure</strong></p><p>The problem states 'first term 1/2' but the sequences given have 1/2 as the <em>second</em> term. We must interpret this carefully: we're looking for APs where 1/2 appears as the second term and the first term satisfies the AP property.</p><p><strong>Step 2: Analyze first sequence: -1/2, 1/2, 3/2, ...</strong></p><p>First term: a₁ = -1/2</p><p>Common difference: d = 1/2 - (-1/2) = 1</p><p>Verify: -1/2, -1/2+1=1/2, 1/2+1=3/2 ✓</p><p>This IS a valid AP with second term = 1/2</p><p><strong>Step 3: Analyze second sequence: -3/2, 1/2, 5/2, ...</strong></p><p>First term: a₁ = -3/2</p><p>Common difference: d = 1/2 - (-3/2) = 2</p><p>Verify: -3/2, -3/2+2=1/2, 1/2+2=5/2 ✓</p><p>This IS a valid AP with second term = 1/2</p><p><strong>Step 4: General insight</strong></p><p>Both sequences are valid APs where a₂ = 1/2. For any AP with second term 1/2: if first term is a₁, then d = 1/2 - a₁, and infinite such progressions exist by varying a₁.</p><p>∴ Answer: A, B (both given sequences correctly exemplify APs with second term 1/2)</p>
Correct Answer: A,B