JEE MAIN - Mathematics (2007 - No. 20)
In the binomial expansion of $${\left( {a - b} \right)^n},\,\,\,n \ge 5,$$ the sum of $${5^{th}}$$ and $${6^{th}}$$ terms is zero, then $$a/b$$ equals
$${{n - 5} \over 6}$$
$${{n - 4} \over 5}$$
$${5 \over {n - 4}}$$
$${6 \over {n - 5}}$$
Explanation
According to the question,
t5 + t6 = 0
$$\therefore$$ $${}^n{C_4}.{a^{n - 4}}.{b^4}$$ + $$\left( { - {}^n{C_5}.{a^{n - 5}}.{b^5}} \right)$$ = 0
By solving we get,
$${a \over b} = {{n - 4} \over 5}$$
t5 + t6 = 0
$$\therefore$$ $${}^n{C_4}.{a^{n - 4}}.{b^4}$$ + $$\left( { - {}^n{C_5}.{a^{n - 5}}.{b^5}} \right)$$ = 0
By solving we get,
$${a \over b} = {{n - 4} \over 5}$$
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