Question

В убывающей арифметической прогрессии a3+a7=6 и a3×a7=8
Найти S16

Answer 1

$\begin{cases}a_3+a_7=6\\a_3\cdot a_7=8\end{cases}\Rightarrow\begin{cases}a_1+2d+a_1+6d=6\\(a_1+2d)(a_1+6d)=8\end{cases}\Rightarrow\\\Rightarrow\begin{cases}2a_1+8d=6\\a_1^2+8a_1d+12d^2=8\end{cases}\Rightarrow\begin{cases}a_1=3-4d\\(3-4d)^2+8d(3-4d)+12d^2=8\end{cases}\\(3-4d)^2+8d(3-4d)+12d^2=8\\9-24d+16d^2+24d-32d^2+12d^2=8\\9-4d^2=8\\4d^2=1\\d^2=\frac14\\d=-\frac12\\\begin{cases}a_1=5\\d=-\frac12\end{cases}\\a_{16}=a_1+15d=5-7,5=-2,5\\S_{16}=\frac{a_1+a_{16}}2\cdot16=\frac{5-2,5}2\cdot16=2,5\cdot8=20$