In the binomial expansion of (a - b)^n, n ≥ 5 , the sum of 5^th and 6^th terms is zero, then a/b equals: - Mathematics MCQ AIEEE 2007 2026 | ExamTest.live

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In the binomial expansion of (a - b)^{n}, n \geq 5, the sum of 5^{t h} and 6^{t h} terms is zero, then \frac{a}{b} equals:

Solution & Step-by-step Explanation

The general term in the expansion of (x + y)^{n} is T_{r + 1} =^{n} C_{r} x^{n - r} y^{r} .For (a - b)^{n} : 5^{t h} term (r = 4): T_{5} =^{n} C_{4} a^{n - 4} (- b)^{4} =^{n} C_{4} a^{n - 4} b^{4} 6^{t h} term (r = 5): T_{6} =^{n} C_{5} a^{n - 5} (- b)^{5} = -^{n} C_{5} a^{n - 5} b^{5} Given T_{5} + T_{6} = 0 :^{n} C_{4} a^{n - 4} b^{4} -^{n} C_{5} a^{n - 5} b^{5} = 0^{n} C_{4} a^{n - 4} b^{4} =^{n} C_{5} a^{n - 5} b^{5} \frac{a ^{n - 4}}{a ^{n - 5}} = \frac{^{n} C _{5} b ^{5}}{^{n} C _{4} b ^{4}} a = \frac{^{n} C _{5}}{^{n} C _{4}} b \frac{a}{b} = \frac{\frac{n !}{5 ! ( n - 5 )!}}{\frac{n !}{4 ! ( n - 4 )!}} = \frac{4 ! ( n - 4 )!}{5 ! ( n - 5 )!} \frac{a}{b} = \frac{4 ! ( n - 4 ) ( n - 5 )!}{5 \times 4 ! ( n - 5 )!} = \frac{n - 4}{5}

Try it yourself before checking the explanation above.

In the binomial expansion of (a - b)^{n}, n \geq 5, the sum of 5^{t h} and 6^{t h} terms is zero, then \frac{a}{b} equals:

\frac{n - 4}{5}

\frac{n - 5}{6}

\frac{5}{n - 4}

\frac{6}{n - 5}

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