Question

9. Hallar la suma de las cifras de "x" en:
40+41+42+…+x=3498

10. Hallar "y" en:
122+132+142+…+y2=15 700

Answer 1

          $3498=40+41+42+\cdots +x\\ \\ 3498 = 1+2+\cdots+39 +40+41+42+\cdots +x-(1+2+\cdots+39)\\ \\ 3498 =\dfrac{x(x+1)}{2}-\dfrac{39(40)}{2}\\ \\ 3498 =\dfrac{x(x+1)}{2}-780\\ \\ \dfrac{x(x+1)}{2} = 4278\\ \\ x^2+x=8556\\ \\ x^2+x-8556=0\\ \\ (x - 92)(x + 93)=0\\ \\ \boxed{x=92}$