If the coefficients of x^9, x^{10} and x^{11} | vedclass

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(C) Given that the coefficients of $x^9, x^{10}$ and $x^{11}$ in the expansion of $(1+x)^n$ are in $A$.$P$.This means ${}^nC_9, {}^nC_{10}$ and ${}^nC

Vedclass · Accepted Answer

$-398$

Vedclass · Answer

(C) Given that the coefficients of $x^9, x^{10}$ and $x^{11}$ in the expansion of $(1+x)^n$ are in $A$.$P$.This means ${}^nC_9, {}^nC_{10}$ and ${}^nC_{11}$ are in $A$.$P$.Therefore,$2({}^nC_{10}) = {}^nC_9 + {}^nC_{11}$.Dividing by ${}^nC_{10}$,we get $2 = \frac{{}^nC_9}{{}^nC_{10}} + \frac{{}^nC_{11}}{{}^nC_{10}}$.Using the formula $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r}$,we have:$2 = \frac{10}{n-9} + \frac{n-10}{11}$.$2 = \frac{110 + (n-10)(n-9)}{11(n-9)}$.$22(n-9) = 110 + n^2 - 19n + 90$.$22n - 198 = n^2 - 19n + 200$.$n^2 - 41n = -198 - 200 = -398$.

If the coefficients of $x^9, x^{10}$ and $x^{11}$ in the expansion of $(1+x)^n$ are in arithmetic progression,then $n^2-41n$ is equal to

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