If the coefficients of $2^{n d}, 3^{r d}, 4^{t h}$ terms of $(1 + x)^{n}$ are in A.P., then $n =$

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Solution

The correct option is D 7
In the expansion $(1 + x)^{n}$
2nd term, $T_{2} =^{n} C_{2} x$ ,
3rd term, $T_{3} =^{n} C_{2} \cdot x^{2}$ ,
4th term, $T_{4} =^{n} C_{3} x^{3}$
Given that coefficients of these are in A.P. $\Rightarrow 2^{n} C_{2} =^{n} C_{1} +^{n} C_{3}$
$\Rightarrow \frac{2 n (n - 1)}{2!} = n + \frac{n (n - 1) (n - 2)}{3!}$
$\Rightarrow n - 1 = 1 + \frac{(n - 1) (n - 2)}{6}$
$\Rightarrow 6 (n - 1) = 6 + n^{2} - 3 n + 2$
$\Rightarrow n^{2} - 9 n + 14 = 0 \Rightarrow n = 7$

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If the coefficients of $2^{n d}$ , $3^{r d}$ and $4^{t h}$ terms in the expansion of $(1 + x)^{n}, n \in N$ are in A.P., then n =

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