\[ a_{45} = a_{21} + (45 - 21)d \]
\[ -158 = -62 + 24d \]
\[ 24d = -158 + 62 \]
\[ 24d = -96 \]
\[ d = -4 \]
\[ a_{21} = a_1 + (21 - 1)d \]
\[ -62 = a_1 + 20(-4) \]
\[ -62 = a_1 - 80 \]
\[ a_1 = -62 + 80 \]
\[ a_1 = 18 \]
Сначала найдем 10-й и 50-й члены:
\[ a_{10} = a_1 + (10 - 1)d = 18 + 9(-4) = 18 - 36 = -18 \]
\[ a_{50} = a_1 + (50 - 1)d = 18 + 49(-4) = 18 - 196 = -178 \]
Сумма с 10-го по 50-й член:
\[ S = \frac{a_{10} + a_{50}}{2} \cdot (50 - 10 + 1) \]
\[ S = \frac{-18 + (-178)}{2} \cdot 41 \]
\[ S = \frac{-196}{2} \cdot 41 \]
\[ S = -98 \cdot 41 \]
\[ S = -4018 \]
Ответ: a_1 = 18, d = -4, S = -4018