11th Class Mathematics Other Series Question Bank Critical Thinking

  • question_answer If  \[x,\,y,z\] are three consecutive positive integers, then \[\frac{1}{2}{{\log }_{e}}x+\frac{1}{2}{{\log }_{e}}z+\frac{1}{2xz+1}+\frac{1}{3}{{\left( \frac{1}{2xz+1} \right)}^{3}}+....=\]

    A) \[{{\log }_{e}}x\]

    B) \[{{\log }_{e}}y\]

    C) \[{{\log }_{e}}z\]

    D) None of these

    Correct Answer: B

    Solution :

    Since \[x,y,z\]are three consecutive positive integers,  therefore \[2y=x+z\]. \[\Rightarrow \,\,4{{y}^{2}}={{(x+z)}^{2}}\Rightarrow 4{{y}^{2}}={{(x-z)}^{2}}+4xz\] \[\Rightarrow \,\,4{{y}^{2}}={{(-2)}^{2}}+4xz,\,\,\,(\because z-x=-2)\] \[\Rightarrow \,\,{{y}^{2}}=1+xz\]                    ....(i)      Now \[\frac{1}{2}{{\log }_{e}}x+\frac{1}{2}{{\log }_{e}}z+\frac{1}{1+2xz}+\frac{1}{3}{{\left( \frac{1}{1+2xz} \right)}^{3}}+....\] \[=\frac{1}{2}\left[ {{\log }_{e}}x+{{\log }_{e}}z \right.\]\[+2\left. \left\{ \left( \frac{1}{1+2xz} \right)+\frac{1}{3}{{\left( \frac{1}{1+2xz} \right)}^{3}}+.... \right\} \right]\] \[=\frac{1}{2}\left[ {{\log }_{e}}xz+{{\log }_{e}}\left( \frac{1+\frac{1}{1+2xz}}{1-\frac{1}{1+2xz}} \right) \right]\] \[=\frac{1}{2}\left[ {{\log }_{e}}xz+{{\log }_{e}}\left( \frac{1+xz}{xz} \right) \right]\] \[=\frac{1}{2}{{\log }_{e}}(1+xz)=\frac{1}{2}{{\log }_{e}}{{y}^{2}}\]\[={{\log }_{e}}y\].


You need to login to perform this action.
You will be redirected in 3 sec spinner