In (a-b)ⁿ (n 5), the sum of the 5th and 6th terms is zero. Then ab equals (a) n-5 6 (b) n-4 5 (c) 5 n-4 (d) 6 n-5

Correct answer: (b) n-4 5 n4aⁿ⁻⁴b⁴= n5aⁿ⁻⁵b⁵→ ab=( n5)/( n4)= n-4 5. JEE Advanced 2001 · Binomial Theorem — verified solution by the Vidaara Team.

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[JEE Advanced 2001] in a b n n 5 the sum of the 5th and 6th terms

VAVidaara Admin Asked 2d ago 0 views 1 answer

In $(a-b)^n$ ($n\ge5$), the sum of the 5th and 6th terms is zero. Then $\dfrac ab$ equals

(a) $\dfrac{n-5}6$
(b) $\dfrac{n-4}5$
(c) $\dfrac5{n-4}$
(d) $\dfrac6{n-5}$

VAVidaara Admin ✓ Vidaara Team ✓ Accepted · 2d ago ▲ 0

Correct answer: (b) $\dfrac{n-4}5$

$\binom n4a^{n-4}b^4=\binom n5a^{n-5}b^5\Rightarrow\frac ab=\frac{\binom n5}{\binom n4}=\frac{n-4}5$.

JEE Advanced 2001 · Binomial Theorem — verified solution by the Vidaara Team.

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