A) 11
B) 10
C) 12
D) 8
Correct Answer: A
Solution :
Let the coefficient of three consecutive terms i.e. \[{{(r+1)}^{th}},\,\,\,\,{{(r+2)}^{th}},\,\,\,{{(r+3)}^{th}}\] in expansion of \[{{(1+x)}^{n}}\] are 165,330 and 462 respectively then, coefficient of \[{{(r+1)}^{th}}\] term \[={}^{n}{{C}_{r}}=165\] Coefficient of (r + 2)th term \[={}^{n}{{C}_{r+1}}=330\] and Coefficient of (r + 3)th term \[={}^{n}{{C}_{r+2}}=462\] \ \[\frac{{}^{n}{{C}_{r+1}}}{{}^{n}{{C}_{r}}}=\frac{n-r}{r+1}=2\] or \[n-r=2(r+1)\] or \[r=\frac{1}{3}(n-2)\] and \[\frac{{}^{n}{{C}_{r+2}}}{{}^{n}{{C}_{r+1}}}=\frac{n-r-1}{r+2}=\frac{231}{165}\] or \[165(n-r-1)=231(r+2)\] or \[165n-627=396r\] or \[165n-627=396\times \frac{1}{3}\times (n-2)\] or \[165n-627=132(n-2)\] or n = 11.
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