Question

The sum of five consecutive positive integers is 55. The sum of the squares of the extreme term is?

Answer 1

Let the integers be 
x
x+1
x+2
x+3
x+4.... x+9 

Thus, A = 5x+10

sum of last 5 terms = 5x + 35 = 5x + 10 +25 = A+25

Answer 2

Let the five consecutive positive integers be x,x+1,x+2,x+3,x+4.

Given, (x) + (x+1) + (x+2)+(x+3)+(x+4) = 55

                 5x + 10 = 55

                      5x = 45

                        x = 9.

The Numbers are 9,10,11,12,13.

The sum of squares of the extreme term is = 9^2 + 13^2 

                                                                         = 81 + 169

                                                                         = 250.


Hope this helps!