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Statement 1:Let p subscript 1 end subscript comma p subscript 2 end subscript comma horizontal ellipsis comma p subscript n end subscript and x be distinct real number such that open parentheses not stretchy sum from r equals 1 to n minus 1 of p subscript r end subscript superscript 2 end superscript close parentheses x to the power of 2 end exponent plus 2 open parentheses not stretchy sum from r equals 1 to n minus 1 of p subscript r end subscript blank p subscript r plus 1 end subscript close parentheses x plus not stretchy sum from r equals 2 to n of p subscript r end subscript superscript 2 end superscript less or equal than 0, then p subscript 1 end subscript comma p subscript 2 end subscript comma horizontal ellipsis comma p subscript n end subscript are in G.P. and when a subscript 1 end subscript superscript 2 end superscript plus a subscript 2 end subscript superscript 2 end superscript plus a subscript 3 end subscript superscript 2 end superscript plus horizontal ellipsis plus a subscript n end subscript superscript 2 end superscript equals 0 comma blank a subscript 1 end subscript equals a subscript 2 end subscript equals a subscript 3 end subscript equals horizontal ellipsis equals a subscript n end subscript equals 0
Statement 2:If fraction numerator p subscript 2 end subscript over denominator p subscript 1 end subscript end fraction equals fraction numerator p subscript 3 end subscript over denominator p subscript 2 end subscript end fraction equals horizontal ellipsis equals fraction numerator p subscript n end subscript over denominator p subscript n minus 1 end subscript end fraction, then p subscript 1 end subscript comma p subscript 2 end subscript comma horizontal ellipsis comma p subscript n end subscript are in G.P.

  1. Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1  
  2. Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1  
  3. Statement 1 is True, Statement 2 is False  
  4. Statement 1 is False, Statement 2 is True  

The correct answer is: Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1


    The given inequality is
    open parentheses p subscript 1 end subscript superscript 2 end superscript plus p subscript 2 end subscript superscript 2 end superscript plus... plus p subscript n minus 1 end subscript superscript 2 end superscript close parentheses x to the power of 2 end exponent plus 2 open parentheses p subscript 1 end subscript p subscript 2 end subscript plus p subscript 2 end subscript p subscript 3 end subscript plus... p subscript n minus 1 end subscript p subscript n end subscript close parentheses x plus open parentheses p subscript 2 end subscript superscript 2 end superscript plus... plus p subscript n end subscript superscript 2 end superscript close parentheses less or equal than 0
    rightwards double arrow blank open parentheses p subscript 1 end subscript x plus p subscript 2 end subscript close parentheses to the power of 2 end exponent plus open parentheses p subscript 2 end subscript x plus p subscript 3 end subscript close parentheses to the power of 2 end exponent plus... plus open parentheses p subscript n minus 1 end subscript x plus p subscript n end subscript close parentheses to the power of 2 end exponent less or equal than 0 (1)
    But each one of the terms on the L.H.S. is a perfect square and hence is positive or zero
    Therefore (1) holds only if
    p subscript 1 end subscript x plus p subscript 2 end subscript equals 0 equals p subscript 2 end subscript x plus p subscript 3 end subscript equals p subscript 3 end subscript x plus p subscript 4 end subscript equals horizontal ellipsis equals p subscript n minus 1 end subscript x plus p subscript n end subscript
    rightwards double arrow blank minus x equals fraction numerator p subscript 2 end subscript over denominator p subscript 1 end subscript end fraction equals fraction numerator p subscript 3 end subscript over denominator p subscript 2 end subscript end fraction equals horizontal ellipsis equals fraction numerator p subscript n end subscript over denominator p subscript n minus 1 end subscript end fraction
    Hence, p subscript 1 end subscript comma p subscript 2 end subscript comma horizontal ellipsis comma p subscript n end subscript are in G.P.

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    Statement 1:Sum of the series 1 to the power of 3 end exponent minus 2 to the power of 3 end exponent plus 3 to the power of 3 end exponent minus 4 to the power of 3 end exponent plus horizontal ellipsis plus 11 to the power of 3 end exponent equals 378
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