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If coefficients of T_9,\;T_{10} \text{ and } T_{11} in the expansion of (1 + x)n are in A.P. then n may be equal to

Option: 1

14


Option: 2

21


Option: 3

20


Option: 4

15


Answers (1)

best_answer

Coefficients of T_{9}=\;^nC_{8},\;T_{10}=\;^nC_{9} \; and \; T_{11}=\;^nC_{10}

As these terms are in A.P., so

2\;^nC_{9} =\;^nC_{8}+\;^nC_{10}

\\2\times\frac{n!}{(n-9)!9!}=\frac{n!}{(n-8)!8!}+\frac{n!}{(n-10)!10!}\\\\\text{Multiplying the equation by }(n-9)!9!\\\\2=\frac{9}{(n-8)}+\frac{n-9}{10}

\begin{array}{c}{20(n-8)=90+n^{2}-17 n+72} \\ {n^{2}-37 n+322=0} \\ {(n-23)(n-14)=0} \\ {n=14,23}\end{array}

hence option A is correct

Posted by

HARSH KANKARIA

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