Question

знайти другий член арифметичної прогресії якщо S20=330 d=-3​

Answer 1

$\displaystyle\bf\\d=-3\\\\S_{20} =330\\\\\\S_{20}=\frac{a_{1} +a _{20} }{2} \cdot 20=(a_{1} + a_{1} +19d)\cdot10=(2a_{1} +19d)\cdot 10\\\\\\(2a_{1} +19d)\cdot 10=330\\\\2a_{1} +19d=33\\\\2a_{1} =33-19d=33-19\cdot (-3)=33+57=90\\\\a_{1} =90:2=45\\\\a_{2} =a_{1} +d=45+(-3)=45-3=42\\\\Otvet \ : \ a_{2} =42$