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JEE Main & Advanced Sample Paper JEE Main Sample Paper-33

  • question_answer
    Let \[{{x}_{n}}\] be the sequence of numbers denoted by\[{{x}_{n}}=\frac{195}{4{{P}_{n}}}-\frac{^{n+3}{{P}_{3}}}{{{P}_{n+1}}}(n\in N)\] where \[{{P}_{n}}\] denotes the number of ways in which n distinct things can be arranged on n different places in a definite order. The sum of all possible values of \[n\in N\] for which \[{{x}_{n}}>0\], is

    A)  10                

    B)                    9

    C)  8         

    D)                                     6

    Correct Answer: A

    Solution :

    We must have,             \[(2n+19)\,(2n-9)\,<0\]             \[\Rightarrow \,\,\frac{-19}{2}<n<\frac{9}{2}\,\Rightarrow \,\,n\in \,\,\{1,\,2,\,3,\,4\}\] Hence, sum \[=1+2+3+4=10\]


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