If the coefficients of x^4, x^5 and x^6 in th | vedclass

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(A) The coefficients of $x^4, x^5, x^6$ in $(1+x)^n$ are $^nC_4, ^nC_5, ^nC_6$ respectively.Since these are in arithmetic progression,we have $2(^nC_5

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$14$

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(A) The coefficients of $x^4, x^5, x^6$ in $(1+x)^n$ are $^nC_4, ^nC_5, ^nC_6$ respectively.Since these are in arithmetic progression,we have $2(^nC_5) = ^nC_4 + ^nC_6$.Dividing by $^nC_5$,we get $2 = \frac{^nC_4}{^nC_5} + \frac{^nC_6}{^nC_5}$.Using the property $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$,we have $\frac{^nC_4}{^nC_5} = \frac{5}{n-4}$ and $\frac{^nC_6}{^nC_5} = \frac{n-5}{6}$.So,$2 = \frac{5}{n-4} + \frac{n-5}{6}$.Multiplying by $6(n-4)$,we get $12(n-4) = 30 + (n-5)(n-4)$.$12n - 48 = 30 + n^2 - 9n + 20$.$n^2 - 21n + 98 = 0$.$(n-14)(n-7) = 0$.Thus,$n = 14$ or $n = 7$.The maximum value of $n$ is $14$.

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

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