<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 numbers \(2A_{171},\ G_5^2 + 1,\ 2A_{172}\) are in</p>
Step-by-Step Solution
Key Concept: Use the properties that arithmetic means between two numbers form an A.P., and geometric means between two numbers form a G.P. The sum formula for A.M.s and product formula for G.M.s help us find the specific terms needed to verify the relationship.
<p><strong>Step 1: Find m (number of arithmetic means)</strong></p><p>The numbers -2, A₁, A₂, ..., Aₘ, 1027 form an A.P. with (m+2) terms.</p><p>Common difference: d = (1027-(-2))/(m+1) = 1029/(m+1)</p><p>Sum of A.M.s: A₁ + A₂ + ... + Aₘ = 1025 × 171</p><p>Using the A.P. property: Sum = m × (first A.M. + last A.M.)/2 = m × [(-2+1027)/2] = m × 512.5</p><p>So: m × 512.5 = 1025 × 171, which gives m × 1025/2 = 1025 × 171</p><p>Therefore: m = 342</p></p><p><strong>Step 2: Find common difference of A.P.</strong></p><p>d = 1029/(342+1) = 1029/343 = 3</p><p>The A.P. is: -2, 1, 4, 7, 10, ..., 1027</p></p><p><strong>Step 3: Find A₁₇₁ and A₁₇₂</strong></p><p>A₁₇₁ = -2 + 171(3) = -2 + 513 = 511</p><p>A₁₇₂ = -2 + 172(3) = -2 + 516 = 514</p><p>Therefore: 2A₁₇₁ = 1022 and 2A₁₇₂ = 1028</p></p><p><strong>Step 4: Find n (number of geometric means)</strong></p><p>The numbers 1, G₁, G₂, ..., Gₙ, 1024 form a G.P. with (n+2) terms.</p><p>Product of G.M.s: G₁ × G₂ × ... × Gₙ = 2⁴⁵</p><p>For a G.P., the product = (G₁ × Gₙ)^(n/2) = [√(1×1024)]ⁿ = 32ⁿ = (2⁵)ⁿ = 2⁵ⁿ</p><p>So: 2⁵ⁿ = 2⁴⁵, which gives 5n = 45, so n = 9</p></p><p><strong>Step 5: Find common ratio and G₅</strong></p><p>For the G.P.: 1, G₁, G₂, ..., G₉, 1024 with (11) terms total</p><p>1024 = 1 × r¹⁰, so r¹⁰ = 2¹⁰, giving r = 2</p><p>G₅ = 1 × 2⁵ = 32</p><p>Therefore: G₅² + 1 = 1024 + 1 = 1025</p></p><p><strong>Step 6: Check if 2A₁₇₁, G₅² + 1, 2A₁₇₂ are in A.P.</strong></p><p>The three terms are: 1022, 1025, 1028</p><p>Differences: 1025 - 1022 = 3 and 1028 - 1025 = 3</p><p>Since consecutive differences are equal, the terms are in A.P.</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A