Sequences & Series
Geometric Progression
Grade 11
Question:
<p>Let \(A_n = \left(\dfrac{3}{4}\right) - \left(\dfrac{3}{4}\right)^2 + \left(\dfrac{3}{4}\right)^3 - \cdots + (-1)^{n-1}\left(\dfrac{3}{4}\right)^n\) and \(B_n = 1 - A_n\). Then the least odd natural number \(p\), so that \(B_n > A_n\) for all \(n \geq p\), is</p>
<p>9</p>
<p>7</p>
<p>11</p>
<p>5</p>
Step-by-Step Solution
Key Concept: Recognize that $A_n$ is a geometric series with first term $a = 3/4$ and common ratio $r = -3/4$. Use the formula for sum of finite geometric series, then find when $B_n = 1 - A_n > A_n$ (i.e., when $A_n < 1/2$).
<p><strong>Step 1:</strong> Find an using the geometric series formula.</p><p>$A_n = \frac{3}{4} \cdot \frac{1-(-3/4)^n}{1-(-3/4)} = \frac{3}{4} \cdot \frac{1-(-3/4)^n}{7/4} = \frac{3}{7}[1-(-3/4)^n]$</p><p><strong>Step 2:</strong> Express the inequality $B_n > A_n$.</p><p>$B_n > A_n \implies 1 - A_n > A_n \implies A_n < \frac{1}{2}$</p><p><strong>Step 3:</strong> Substitute and solve.</p><p>$\frac{3}{7}[1-(-3/4)^n] < \frac{1}{2}$</p><p>$1-(-3/4)^n < \frac{7}{6}$</p><p>$-(-3/4)^n < \frac{1}{6}$</p><p>$(-3/4)^n > -\frac{1}{6}$</p><p><strong>Step 4:</strong> Analyze for odd $n$.</p><p>For odd $n$: $(-3/4)^n = -(3/4)^n < 0$</p><p>We need: $-(3/4)^n > -\frac{1}{6} \implies (3/4)^n < \frac{1}{6}$</p><p>Testing: $n=1$: $(3/4)^1 = 0.75 > 1/6$ ✗</p><p>$n=3$: $(3/4)^3 = 27/64 ≈ 0.422 > 1/6$ ✗</p><p>$n=5$: $(3/4)^5 = 243/1024 ≈ 0.237 > 1/6$ ✗</p><p>$n=7$: $(3/4)^7 = 2187/16384 ≈ 0.133 > 1/6 ≈ 0.167$ ✗</p><p>$n=9$: $(3/4)^9 ≈ 0.075 < 1/6$ ✓</p><p>∴ Answer: <strong>B (p = 9)</strong></p>
Correct Answer: B