A) - 7
B) 7
C) 14
D) - 14
Correct Answer: B
Solution :
Coefficients of 2nd, 3rd and 4th terms are respectively \[^{n}{{C}_{1}},{{\,}^{n}}{{C}_{2}}\] and \[^{n}{{C}_{3}}\] are in A.P. Þ \[{{2.}^{n}}{{C}_{2}}={{\,}^{n}}{{C}_{1}}+{{\,}^{n}}{{C}_{3}}\] Þ \[\frac{2n!}{2\,!\left( n-2 \right)\,!}=\frac{n!}{(n-1)!}+\frac{n!}{3!\,\left( n-3 \right)\,!}\] On solving, \[{{n}^{2}}-9n+14=0\,\,\Rightarrow \,\,{{n}^{2}}-9n=-14\].
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