SOLUTION: The sum of four numbers in an arithmetic progression is 98. The sum of their squares is 3006. Find the third number.

Let  ,  ,  ,    be four terms of the AP.

Let  "c"  be the central point in the number line, exactly half way between the terms  and .

Let d be the  HALF  of the common difference of the progression.


Then   = c - 3d,

       = c -  d,

       = c +  d,

       = c + 3d.


Then the sum of the four terms is 4d,  and it equals 98, so  c = 98/4 = 24.5

The sum of squares of the four terms is

    (c-3d)^2 + (c-d)^2 + (c+d)^2 + (c+3d)^2 = 3006


Making FOIL and combining like terms, you arrive to equation

    4c^2 + 20d^2 = 3006,

    4*24.5^2 + 20d^2 = 3006

    20d^2 = 3006 - 4*24.5^2 = 605

      d^2 = 605/20 = 30.25

      d =  = +/- 5.5.


Thus the four terms of the progression are  24.5-3*5.5 = 8,      24.5-5.5 = 19,     30,  41,  if  d= 5.5,  and

                                            24.5-3*(-5.5) = 41,  24.5-(-5.5) = 30,  19,   8,  if d= -5.5.


Thus the problem has two possible  ANSWERS  for the third term: it is EITHER  30  OR  19.