For some \mathrm{n}
eq 10, let the coefficients of the 5 th, 6 th and 7 th terms in the binomial expansion of (1+\mathrm{x})^{\mathrm{n}+4} be in A.P. Then the largest coefficient in the expansion of (1+\mathrm{x})^{\mathrm{n}+4} is:

Question

For some \mathrm{n} 
eq 10, let the coefficients of the 5 th, 6 th and 7 th terms in the binomial expansion of (1+\mathrm{x})^{\mathrm{n}+4} be in A.P. Then the largest coefficient in the expansion of (1+\mathrm{x})^{\mathrm{n}+4} is:

Accepted Answer

Answer: B. 35 \begin{aligned} & (1+\mathrm{x})^{\mathrm{n+4}} \ & { }^{\mathrm{n}+4} \mathrm{C}_4{ }^{\mathrm{n+4}} \mathrm{C}_5{ }^{\mathrm{n}+4} \mathrm{C}_6, ightarrow 	ext { A.P. } \ & \Rightarrow 2 	imes{ }^{\mathrm{n}+4} \mathrm{C}_5={ }^{\mathrm{n+4}} \mathrm{C}_4+{ }^{\mathrm{n}+4} \mathrm{C}_6 \ & \Rightarrow 4 	imes{ }^{\mathrm{n+4}} \mathrm{C}_5=\left({ }^{n+4} \mathrm{C}_4+{ }^{\mathrm{n}+4} \mathrm{C}_5ight)+\left({ }^{\mathrm{n+4}} \mathrm{C}_5+{ }^{\mathrm{n}+4} \mathrm{C}_6ight) \ & \Rightarrow 4 	imes{ }^{n+4} \mathrm{C}_5={ }^{\mathrm{n}+5} \mathrm{C}_5+{ }^{\mathrm{n}+5} \mathrm{C}_6 \ & \Rightarrow 4 	imes \frac{(\mathrm{n}+4)!}{5!.(\mathrm{n}-1)!}=\frac{(\mathrm{n}+6)!}{6!\mathrm{n}!} \ & \Rightarrow 4=\frac{(\mathrm{n}+6)(\mathrm{n}+5)}{6 \mathrm{n}} \ & \Rightarrow \mathrm{n}^2+11 \mathrm{n}+30=24 \mathrm{n} \ & \Rightarrow \mathrm{n}^2-13 \mathrm{n}+30=0 \ & \Rightarrow \mathrm{n}=3,10(	ext { rejected }) \ & \because \mathrm{n} 
eq 10 \end{aligned} 	herefore Largest binomial coefficient in expansion of \begin{aligned} & (1+x)^7 \ & (\because \mathrm{n}+4=7) \end{aligned} is coeff. of middle term \Rightarrow{ }^7 \mathrm{C}_4={ }^7 \mathrm{C}_3=35 N.T.A. Ans Option (2)

For some $\mathrm{n} \neq 10$, let the coefficients of the 5 th, 6 th and 7 th terms in the binomial expansion of $(1+\mathrm{x})^{\mathrm{n}+4}$ be in A.P. Then the largest coefficient in the expansion of $(1+\mathrm{x})^{\mathrm{n}+4}$ is:

Answer

Explanation

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