Question

The sequence $4 ; 9 ; x ; 37 ; ... $ is a quadratic sequence.

1. Calculate $x$
2. Hence, or otherwise, determine the nth term of the sequence

Answer 1

1. 1st difference= $5,x-9,37-x,...$

    2nd difference= $x-9-5, 37-x-(x-9),...$

since 2nd difference is constant,

  $x-9-5 = 37-x-(x-9)$

 $  x-14=37-2x+9$

  $   3x= 60$

       $     x=20$

2.  $2a= x-9-5 = 20-9-5= 6$

     $2a= 6$

   $ a= 3$

$3a+b=5$

    $   b= 5-3(3)$

   $      b= -4$

$a+b+c= 4$

          $c= 4-3-(-4)$

          $   c= 5$

Therefore nth term = $3n^2 -4n +5$