Sequences & Series
Geometric Progression
Grade 11
Question:
<p><strong>39.</strong> Let \(a = 1\,1\,1\,\ldots1\) (55 digits), \(b = 1 + 10 + 10^2 + \cdots + 10^4\), \(c = 1 + 10^5 + 10^{10} + 10^{15} + \cdots + 10^{50}\), then</p>
<p>\(a = b + c\)</p>
<p>\(a = bc\)</p>
<p>\(b = ac\)</p>
<p>\(c = ab\)</p>
Step-by-Step Solution
Key Concept: Express a as a geometric series: a = (10^55 - 1)/9. Recognize that b = (10^5 - 1)/9 and c = (10^55 - 1)/(10^5 - 1). The key insight is that a = b·c because the repunit with 55 ones can be factored using the divisibility of exponents.
<p><strong>Step 1:</strong> Express each number as a geometric series.</p><p>a = 111...1 (55 digits) = (10^55 - 1)/9</p><p>b = 1 + 10 + 10^2 + 10^3 + 10^4 = (10^5 - 1)/9</p><p><strong>Step 2:</strong> Recognize that c has 11 terms (k = 0, 1, 2, ..., 10):</p><p>c = 1 + 10^5 + 10^10 + ... + 10^50 = (10^55 - 1)/(10^5 - 1)</p><p>This uses the formula for geometric series: 1 + r + r^2 + ... + r^10 = (r^11 - 1)/(r - 1) with r = 10^5</p><p><strong>Step 3:</strong> Compute b·c:</p><p>b·c = [(10^5 - 1)/9] · [(10^55 - 1)/(10^5 - 1)]</p><p>= (10^55 - 1)/9 = a</p><p><strong>Step 4:</strong> Therefore, a = b·c, so the relation is <strong>a = bc</strong></p><p>∴ Answer: B</p>
Correct Answer: B