JEE Main & Advanced JEE Main Paper (Held On 8 April 2017)

  • question_answer
    If the um of the first n terms of the series\[\sqrt{3}+\sqrt{75}+\sqrt{243}+\sqrt{507}+........\]is\[435\sqrt{3},\]then n equals. [JEE Online 08-04-2017]

    A)  13                                         

    B)  15

    C)  29                                         

    D)  18

    Correct Answer: B

    Solution :

                    \[\sqrt{3}[1+\sqrt{25}+\sqrt{81}+\sqrt{69}+......]=435\sqrt{3}\]                 \[\sqrt{3}[1+5+9+13+.....{{T}_{n}}]=435\sqrt{3}\]                 \[=\sqrt{3}\times \frac{n}{2}[2+(n-1)4]=435\sqrt{3}\]                 \[2n+4{{n}^{2}}-4n=870\]                 \[=4{{n}^{2}}-2n-870=0\]                 \[=2{{n}^{2}}-n-435=0\]                 \[n=\frac{1\pm \sqrt{1+4\times 2\times 435}}{4}\]                 \[=\frac{1\pm 59}{4}\]                 \[=\frac{1+59}{4}=4;\frac{1-59}{4}\]


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