SOLUTION: Three numbers are in AP such that their sum is 18 and sum of their squares is 158 the greatest number among them is

Let x be the middle term of our 3-term AP.

Then  = x-d,  = x and  = x + d, and the sum of the three terms is  (x-d) + x + (x+d) = 3d.

So we have 3x = 18 and x =  = 6.


Next, the square of the first term is  = ,
      the square of the middle term is  and
      the square of the third term is  = .

Add these tree squares, and you will get their sum as .
Now recall that x = 6, hence, x^2 = 36, and the equation for the squares becomes

 = 158.

--->   = 158-108 = 50  --->   =  = 25 --->  d = +/- 5.


So, there are two AP progressions satisfying the condition:

1)  6-5 = 1, 6,  6+5 = 11,   and  the reversed sequence


2)  11, 6, 5.