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JEE Main & Advanced Mathematics Sequence & Series Question Bank Critical Thinking

  • question_answer
    The first term of an infinite geometric progression is \[x\] and its sum is 5. Then [IIT Screening 2004]

    A) \[0\le x\le 10\]

    B) \[0<x<10\]

    C) \[-10<x<0\]

    D) \[x>10\]

    Correct Answer: B

    Solution :

      \[5=\frac{x}{1-r}\] \[\Rightarrow \] \[5-5r=x\] \[\Rightarrow \]  \[r=1-\frac{x}{5}\] As \[|r|\,<\,1\]  i.e.,  \[\left| \,1-\frac{x}{5}\, \right|<1\] \[-1<1-\frac{x}{5}<1\] \[-5<5-x<5=-10<-x<0=10>x>0\] i.e,  \[0<x<10\].


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