Show that
$open vertical bar table row cell straight x plus 1 end cell cell space straight x plus 2 end cell cell space space straight x plus straight a end cell row cell straight x plus 2 end cell cell space space space straight x plus 3 space space end cell cell straight x plus straight b end cell row cell straight x plus 3 end cell cell space straight x plus 4 end cell cell space straight x plus straight c end cell end table close vertical bar space equals 0$ ,where a,b,c and in A.P

Solution

Since space straight a comma space straight b comma space straight c space are space in space straight A. straight P. space space space therefore space 2 space straight b space equals space straight a space plus space straight c space space space space space space space space space space space space... left parenthesis 1 right parenthesis straight L. straight H. straight S space space space equals space open vertical bar table row cell straight x plus 1 end cell cell space straight x plus 2 end cell cell space space straight x plus straight a end cell row cell straight x plus 2 end cell cell space space space straight x plus 3 space space end cell cell straight x plus straight b end cell row cell straight x plus 3 end cell cell space straight x plus 4 end cell cell space straight x plus straight c end cell end table close vertical bar space space space space space space space space space space space space space space equals space open vertical bar table row cell straight x plus 1 end cell cell space straight x plus 2 end cell cell space space straight x plus straight a end cell row cell 2 straight x plus 4 end cell cell space space 2 space straight x plus 6 space end cell cell 2 straight x plus straight c plus straight a end cell row cell straight x plus 3 end cell cell space straight x plus 4 end cell cell space straight x plus straight c end cell end table close vertical bar space space comma space multiplying space straight R subscript 2 space by space 2 space space space space space space space space space space space space space space space space space space equals space space open vertical bar table row cell straight x plus 1 end cell cell space straight x plus 2 end cell cell space space straight x plus straight a end cell row cell 2 straight x plus 4 end cell cell space space 2 space straight x plus 6 space end cell cell 2 straight x plus straight c plus straight a end cell row cell straight x plus 3 end cell cell space straight x plus 4 end cell cell space straight x plus straight c end cell end table close vertical bar space space space space open square brackets therefore table row cell of space left parenthesis 1 right parenthesis end cell end table close square brackets space space space space space space space space space space space space space space equals space space open vertical bar table row cell straight x plus 1 end cell cell space space space straight x plus 2 end cell cell space space straight x plus straight a end cell row 0 cell space space 0 space end cell 0 row cell straight x plus 3 end cell cell space straight x plus 4 end cell cell space space straight x plus straight c space space end cell end table space close vertical bar space space space comma space by space straight R subscript 2 minus straight R subscript 1 minus straight R subscript 3 space space space space space space space space space space space space space equals space 0 space space space space space space space space space space space space space open square brackets table row cell because space each space element space of space straight R subscript 2 end cell end table space is space zero close square brackets space space space space space space space space space space space space space equals space space space straight R. straight H. straight S space space

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