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JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Afternoon)

  • question_answer
                If some three consecutive in the binomial expansion of \[{{\left( x+1 \right)}^{n}}\]is powers of x are in the ratio 2 : 15 : 70, then the average of these three coefficient is :-             [JEE Main 9-4-2019 Afternoon]

    A) 964                 

    B) 625

    C) 227                 

    D) 232

    Correct Answer: D

    Solution :

    \[\frac{^{n}{{C}_{r-1}}}{^{n}{{C}_{r}}}=\frac{2}{15}\]           \[\frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}=\frac{2}{15}\]           \[\frac{r}{n-r+1}=\frac{2}{15}\]           \[15r=2n-2r+2\]           \[\]           \[\frac{^{n}{{C}_{r}}}{^{n}{{C}_{r+1}}}=\frac{15}{70}\]           \[\frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r+1)!(n-r-1)!}}=\frac{3}{14}\]           \[\frac{r+1}{n-r}=\frac{3}{14}\]           \[14r+14=3n-3r\]                     \[\begin{align}   & 3n-17r=14 \\  & \frac{2n-17r=-2}{n=16} \\ \end{align}\]           \[17r=34,r=2\]           \[^{16}{{C}_{1}}{{,}^{16}}{{C}_{2}}{{,}^{16}}{{C}_{3}}\] \[\frac{^{16}{{C}_{1}}{{+}^{16}}{{C}_{2}}{{+}^{16}}{{C}_{3}}}{3}=\frac{16+120+560}{3}\]           \[\frac{680+16}{3}=\frac{696}{3}=232\]          


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