Решите уравнение: 3+7+11+…+x=136, где x – натуральное число.

\[3 + 7 + 11\ldots + x = 136\]

\[a_{1} = 3;\ \ d = 4;\ \ \]

\[a_{n} = 3 + 4 \cdot (n - 1) =\]

\[= 3 + 4n - 4 = 4n - 1\]

\[S = \frac{3 + 4n - 1}{2} \cdot n =\]

\[= \frac{4n + 2}{2} \cdot n = (2n + 1)n =\]

\[= 2n^{2} + n = 136\]

\[2n^{2} + n - 136 = 0\ \ \]

\[D = 1 + 1088 = 1089\]

\[n = \frac{- 1 - 33}{4} < 0\ \ \ \ \]

\[n = \frac{- 1 + 33}{4} = 8\]

\[x = 4 \cdot 8 - 1 = 31.\]

\[Ответ:x = 31.\]