Решите уравнение: 4+10+16+…+x=310, где x – натуральное число.

\[4 + 10 + 16 + \ldots + x = 310\]

\[a_{1} = 4,\ \ d = 6\]

\[\frac{2a_{1} + d(n - 1)}{2} \cdot n = 310\]

\[\frac{4 + x}{2} \cdot n = 310\ \ \ \]

\[a_{n} = 6n - 2\ \ \ \ \]

\[x = 6n - 2\]

\[\frac{6n + 2}{2} \cdot n = 310\]

\[(3n + 1)n = 310\]

\[3n^{2} + n - 310 = 0\]

\[D = 1 + 3720 = 3721\ \ \ \ \]

\[\ n_{1} = \frac{- 1 - 61}{6} \Longrightarrow не\ \]

\[удовлетворяет\]

\[n_{2} = \frac{- 1 + 61}{6} = 10\ \ \ \ \]

\[x = 6 \cdot 10 - 2 = 58\]

\[Ответ:x = 58.\]