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Q.
If the coefficients of $2^{nd}$, $3^{rd}$ and $4^{th}$ terms in the expansion of $(1 + x)^n$are in A.P., then value of $n$ is
Binomial Theorem
Solution:
Given expansion is $(1 + x)^n$.
Coefficients of $2^{nd}$, $3^{rd}$ and $4^{th}$ terms are $^{n}C_{1}$, $^{n}C_{2}$ and $^{n}C_{3}$ respectively.
Since, $^{n}C_{1}$, $^{n}C_{2}$, $^{n}C_{3}$ are in A.P.
$\Rightarrow 2^{n}C_{2}= \,{}^{n}C_{1}+\,{}^{n}C_{3}$
$\Rightarrow 2=\frac{^{n}C_{1}}{^{n}C_{2}}+\frac{^{n}C_{3}}{^{n}C_{2}}$
$\Rightarrow 2= \frac{2}{n-1}+\frac{n-2}{3}$
$\Rightarrow n^{2} - 9n + 14 = 0$
$\Rightarrow n = 2$, $7$
But $n \ne 2$
$\therefore n = 7$