Sequences & Series
Geometric Progression
Grade 11
Question:
<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>If the decreasing G.P. is considered, then the sum of infinite terms is</p>
<p>64</p>
<p>128</p>
<p>256</p>
<p>729</p>
Step-by-Step Solution
Key Concept: For a decreasing G.P., identify the common ratio |r| < 1, then use S∞ = a/(1-r) where 'a' is the first term. The constraints given determine both 'a' and 'r' uniquely.
<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 given conditions:</p><ul><li>First + Last: <em>a</em> + ar^(n-1) = 66 ... (1)</li><li>Second × (Last-1): ar · ar^(n-2) = a²r^(n-1) = 128 ... (2)</li><li>Sum of terms: a(r^n - 1)/(r - 1) = 126 ... (3)</li></ul><p><strong>Step 3:</strong> From (1) and (2): a² r^(n-1) = 128 and a + ar^(n-1) = 66</p><p>Let ar^(n-1) = L (last term). Then: aL = 128 and a + L = 66</p><p>This gives L² - 66L + 128 = 0 → L = 64 or L = 2</p><p><strong>Step 4:</strong> For decreasing G.P., we need a > L, so a = 64 and L = 2</p><p>Therefore: ar^(n-1) = 2 and a = 64 → r^(n-1) = 1/32</p><p><strong>Step 5:</strong> From (3): 64(r^n - 1)/(r - 1) = 126</p><p>Since r^(n-1) = 1/32, we have r^n = r/32</p><p>64(r/32 - 1)/(r - 1) = 126 → 64r/32 - 64)/(r - 1) = 126</p><p>2(r - 32)/(r - 1) = 126 → 2r - 64 = 126r - 126 → r = 1/2</p><p><strong>Step 6:</strong> For infinite G.P. with a = 64 and r = 1/2:</p><p>S∞ = a/(1 - r) = 64/(1 - 1/2) = 64/(1/2) = <strong>128</strong></p><p>∴ Answer: C</p>
Correct Answer: C