Question

sa se gaseasca 101 numere intregi consecutive a caror suma este egala cu 5050

Answer 1

a+(a+1)+.....(a+100)=5050
ax101+ 1+2+3+.....+100=5050
ax101+ 101x100/2=5050
ax101+101x50=5050
ax101+5050=5050
ax101=0
a=0
numerele vor fi 0,1,2,3,.....,101.

Answer 2

(n - 50) + (n-49) + .........+(n-1) +n + (n+1) +.........+(n + 49) + (n + 50) = 5050
101n = 5050             n = 50
0,1,2,3,4............99,100