\[\begin{array}{l}1 + 7 + 13 + ...... + x = 280\\ \Leftrightarrow \left( {\frac{{x - 1}}{6} + 1} \right).\frac{{1 + x}}{2} = 280\\ \Leftrightarrow \frac{{\left( {x - 1 + 6} \right)}}{6}.\frac{{\left( {x + 1} \right)}}{2} = 280\\ \Leftrightarrow \left( {x + 5} \right).\left( {x + 1} \right) = 3360\\ \Leftrightarrow {x^2} + 6x + 5 - 3360 = 0\\ \Leftrightarrow {x^2} + 6x - 3355 = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 55\\x = - 61\,\,\left( {loai} \right)\end{array} \right.\end{array}\]
Vậy \(x = 55\)
