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    Maths-
    General
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    Question

    If x comma y comma zare three consecutive positive integers, then fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application x plus fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application z plus fraction numerator 1 over denominator 2 x z plus 1 end fraction plus fraction numerator 1 over denominator 3 end fraction open parentheses fraction numerator 1 over denominator 2 x z plus 1 end fraction close parentheses to the power of 3 end exponent plus.... equals

    1. log subscript e end subscript invisible function application x    
    2. log subscript e end subscript invisible function application y    
    3. log subscript e end subscript invisible function application z    
    4. None of these    

    The correct answer is: log subscript e end subscript invisible function application y

      Since x comma y comma zare three consecutive positive integers, therefore 2 y equals x plus z.
      rightwards double arrow 4 y to the power of 2 end exponent equals left parenthesis x plus z right parenthesis to the power of 2 end exponent rightwards double arrow 4 y to the power of 2 end exponent equals left parenthesis x minus z right parenthesis to the power of 2 end exponent plus 4 x z
      rightwards double arrow 4 y to the power of 2 end exponent equals left parenthesis negative 2 right parenthesis to the power of 2 end exponent plus 4 x z comma left parenthesis because z minus x equals negative 2 right parenthesis
      rightwards double arrow y to the power of 2 end exponent equals 1 plus x z ....(i)
      Now fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application x plus fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application z plus fraction numerator 1 over denominator 1 plus 2 x z end fraction plus fraction numerator 1 over denominator 3 end fraction open parentheses fraction numerator 1 over denominator 1 plus 2 x z end fraction close parentheses to the power of 3 end exponent plus....
      equals fraction numerator 1 over denominator 2 end fraction open square brackets log subscript e end subscript invisible function application x plus log subscript e end subscript invisible function application z close plus 2 open open curly brackets open parentheses fraction numerator 1 over denominator 1 plus 2 x z end fraction close parentheses plus fraction numerator 1 over denominator 3 end fraction open parentheses fraction numerator 1 over denominator 1 plus 2 x z end fraction close parentheses to the power of 3 end exponent plus.... close curly brackets close square brackets
      equals fraction numerator 1 over denominator 2 end fraction open square brackets log subscript e end subscript invisible function application x z plus log subscript e end subscript invisible function application open parentheses fraction numerator 1 plus fraction numerator 1 over denominator 1 plus 2 x z end fraction over denominator 1 minus fraction numerator 1 over denominator 1 plus 2 x z end fraction end fraction close parentheses close square brackets
      equals fraction numerator 1 over denominator 2 end fraction open square brackets log subscript e end subscript invisible function application x z plus log subscript e end subscript invisible function application open parentheses fraction numerator 1 plus x z over denominator x z end fraction close parentheses close square brackets
      equals fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application left parenthesis 1 plus x z right parenthesis equals fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application y to the power of 2 end exponent equals log subscript e end subscript invisible function application y.

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