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: [2024]

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Solution

Q.
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: [2024]

1 14

2 21

3 7

4 28

Ans.
(1)

As ${(1 + x)}^{n} = {}^{n}C_{0} + {}^{n}{C_{1}} x^{1} + {}^{n}{C_{2}} x^{2} + \dots + {}^{n}{C_{n}} x^{n}$

$∴$ ${}^{n}{C_{5}} - {}^{n}{C_{4}} = {}^{n}{C_{6}} - {}^{n}{C_{5}} [Since, {}^{n}{C_{4}}, {}^{n}{C_{5}} and {}^{n}{C_{6}} are in A.P.]$

$\Rightarrow \frac{n!}{5! (n - 5)!} - \frac{n!}{4! (n - 4)!} = \frac{n!}{6! (n - 6)!} - \frac{n!}{5! (n - 5)!}$

$\Rightarrow \frac{n - 9}{5! (n - 5)! (n - 4)} = \frac{n - 11}{6! (n - 6)! (n - 5)}$

$\Rightarrow 6 (n - 9) = (n - 11) (n - 4)$

$\Rightarrow n^{2} - 21 n + 98 = 0 \Rightarrow n = \frac{21 \pm \sqrt{441 - 392}}{2} = 14, 7$

$∴$ $n_{max} = 14$

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