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:

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Coeff. of } x^4={ }^n C_4 \ & 	ext { Coeff. of } x^5={ }^n C_5 \ & 	ext { Coeff. of } x^6={ }^n C_6 \ & { }^n C_4,{ }^n C_5,{ }^n C_6 \ldots . A P \ & 2 C_5={ }^n C_4+{ }^n C_6 \ & 2=\frac{{ }^n C_4}{{ }^n C_5}+\frac{{ }^n C_6}{{ }^n C_5} \quad\left\{\frac{{ }^n C_r}{{ }^n C_r}=\frac{n-r+1}{r}ight\} \ & 2=\frac{5}{n-4}+\frac{n-5}{6} \ & 12(n-4)=30+n^2-9 n+20 \ & n^2-21 n+98=0 \ & (n-14)(n-7)=0 \ & n_{\max }=14 \quad n_{\min }=7\end{aligned}$

If the coefficients of $x^4, x^5$ and $x^6$ in the expansion of $(1+x)^n$ are in the arithmetic progression,…

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