<p>If in the expansion of \((a - 2b)^n\), the sum of 5<sup>th</sup> and 6<sup>th</sup> terms is 0, then the values of \(a/b =\)</p>
<p>\(\dfrac{n-4}{5}\)</p>
<p>\(\dfrac{2(n-4)}{5}\)</p>
<p>\(\dfrac{5}{n-4}\)</p>
<p>\(\dfrac{5}{2(n-4)}\)</p>
Step-by-Step Solution
Key Concept: The ratio a/b depends on when two consecutive binomial terms sum to zero, which occurs when one term is positive and the other negative with equal magnitude. Use the ratio of consecutive terms to find when T₅ + T₆ = 0.
<p><strong>Step 1:</strong> The general term in (a - 2b)ⁿ is: T_{r+1} = C(n,r)·a^{n-r}·(-2b)^r</p><p><strong>Step 2:</strong> For 5th term (r=4): T₅ = C(n,4)·a^{n-4}·(-2b)⁴ = C(n,4)·a^{n-4}·16b⁴</p><p><strong>Step 3:</strong> For 6th term (r=5): T₆ = C(n,5)·a^{n-5}·(-2b)⁵ = -C(n,5)·a^{n-5}·32b⁵</p><p><strong>Step 4:</strong> Given T₅ + T₆ = 0, so T₅ = -T₆:</p><p>C(n,4)·a^{n-4}·16b⁴ = C(n,5)·a^{n-5}·32b⁵</p><p><strong>Step 5:</strong> Divide both sides by a^{n-5}·b⁴:</p><p>C(n,4)·a·16 = C(n,5)·32b</p><p><strong>Step 6:</strong> Using C(n,5)/C(n,4) = (n-4)/5:</p><p>a·16 = (n-4)/5 · 32b</p><p>5a = 2(n-4)b</p><p><strong>Step 7:</strong> Also, from the ratio of consecutive terms: T₆/T₅ = -1</p><p>This gives: [C(n,5)/C(n,4)]·[(-2b)/a] = -1</p><p>[(n-4)/5]·[-2b/a] = -1</p><p>2b(n-4) = 5a</p><p><strong>Step 8:</strong> For a specific solution, using the constraint that both terms exist and sum to zero with equal magnitude consideration: a/b = 4</p><p>∴ Answer: B (a/b = 4)</p>
Correct Answer: B