Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p><strong>For Problems 19–21:</strong> Let \(A_1, A_2, A_3, \ldots, A_m\) be the arithmetic means between \(-2\) and 1027 and \(G_1, G_2, G_3, \ldots, G_n\) be the geometric means between 1 and 1024. The product of geometric means is \(2^{45}\) and sum of arithmetic means is \(1025 \times 171\).</p><p>The number of arithmetic means is</p>
<p>442</p>
<p>342</p>
<p>378</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Use the sum formula for arithmetic means: if there are m AMs between a and b, then their sum equals m times the average of a and b. Similarly, use the product property of geometric means: if there are n GMs between a and b, their product equals (ab)^(n/2).
<p><strong>Step 1:</strong> Use the sum property of arithmetic means. If A₁, A₂, ..., Aₘ are m arithmetic means between -2 and 1027, then these m+2 terms form an AP with first term a = -2 and last term l = 1027.</p><p><strong>Step 2:</strong> The sum of all m arithmetic means is given by: Sum of AMs = m × (first term + last term)/2 = m × (-2 + 1027)/2 = m × 1025/2</p><p><strong>Step 3:</strong> We're given that the sum of arithmetic means equals 1025 × 171. Therefore:</p><p>m × 1025/2 = 1025 × 171</p><p><strong>Step 4:</strong> Divide both sides by 1025:</p><p>m/2 = 171</p><p><strong>Step 5:</strong> Solve for m:</p><p>m = 342</p><p>∴ Answer: B</p>
Correct Answer: B