<p><strong>For Problems 7–9:</strong> In a G.P., the sum of the first and last terms is 66, the product of the second and the last but one is 128, and the sum of the terms is 126.</p><p>In any case, the difference of the least and greatest terms is</p>
Step-by-Step Solution
Key Concept: In a G.P. with first term a, common ratio r, and n terms: use the constraints a + ar^(n-1) = 66, ar · ar^(n-2) = 128 (which gives a²r^(n-1) = 128), and sum formula S_n = a(r^n - 1)/(r - 1) = 126 to find all terms, then compute max - min.
<p><strong>Step 1:</strong> Let the G.P. have first term <em>a</em>, common ratio <em>r</em>, and <em>n</em> terms.</p><p><strong>Step 2:</strong> From the given conditions:</p><p>• a + ar^(n-1) = 66 ... (1)</p><p>• (ar)(ar^(n-2)) = 128 ⟹ a²r^(n-1) = 128 ... (2)</p><p>• Sum S_n = a(r^n - 1)/(r - 1) = 126 ... (3)</p><p><strong>Step 3:</strong> From (1) and (2): Divide (2) by (1):</p><p>a²r^(n-1)/(a + ar^(n-1)) = 128/66 = 64/33</p><p>Let a + ar^(n-1) = 66, then a(a + ar^(n-1)) = 128 · 33/64 = 66a</p><p>This gives: a · 66 = 66a (consistent). Now a²r^(n-1) = 128, so ar^(n-1) = 128/a</p><p><strong>Step 4:</strong> From a + ar^(n-1) = 66: a + 128/a = 66</p><p>⟹ a² - 66a + 128 = 0</p><p>⟹ a = (66 ± √(4356 - 512))/2 = (66 ± √3844)/2 = (66 ± 62)/2</p><p>⟹ a = 64 or a = 2</p><p><strong>Step 5:</strong> If a = 64: ar^(n-1) = 2, and if a = 2: ar^(n-1) = 64</p><p><strong>Step 6:</strong> Testing a = 2, ar^(n-1) = 64 with sum condition: For n = 6, r = 2:</p><p>S_6 = 2(2^6 - 1)/(2-1) = 2(63) = 126 ✓</p><p>Terms: 2, 4, 8, 16, 32, 64</p><p><strong>Step 7:</strong> Maximum term = 64, Minimum term = 2</p><p>∴ Difference = 64 - 2 = <strong>62</strong></p>
Correct Answer: B