JEE MAIN - Mathematics (2024 - 4th April Evening Shift - No. 2)

If the coefficients of $$x^4, x^5$$ and $$x^6$$ in the expansion of $$(1+x)^n$$ are in the arithmetic progression, then the maximum value of $$n$$ is:
28
21
7
14

Explanation

$$\begin{aligned} & (1+x)^n={ }^n C_0+{ }^n C_1 x^1+{ }^n C_2 x^2+\ldots{ }^n C_n x^n \\ & { }^n C_4,{ }^n C_5 \&{ }^n C_6 \text { are in A.P. } \\ & { }^n C_5-{ }^n C_4={ }^n C_6-{ }^n C_5 \\ & \Rightarrow \frac{n!}{5!(n-5)!}-\frac{n!}{4!(n-4)!}=\frac{n!}{6!(n-6)!}-\frac{n!}{5!(n-5)!} \\ & \Rightarrow 30(n-9)(n-6)=5(n-4)(n-11) \\ & \Rightarrow 30 n^2-450 n+1620=5 n^2 \\ & \Rightarrow \frac{1}{n-5}\left[\frac{n-4-5}{5(n-4)}\right]=\frac{1}{5}\left[\frac{n-5-6}{6(n-5)}\right] \\ & \Rightarrow \frac{n-9}{5(n-4)}=\frac{1}{5}\left[\frac{n-11}{6}\right] \\ & \Rightarrow n^2-21 n+98=0 \\ & n_{\max }=14 \end{aligned}$$

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