Đáp án:
\(U_{1}=3\)
d=2
S=50600
Giải thích các bước giải:
a) Ta có:
Đề \( \Leftrightarrow \)\(\left\{\begin{matrix} U_{1}+d+U_{1}+4d-U_{1}-2d=9
& & \\ U_{1}+U_{1}+5d=16
& &
\end{matrix}\right.\)
\( \Leftrightarrow \)\(\left\{\begin{matrix}U_{1}+3d=9
& & \\ 2U_{1}+5d=16
& &
\end{matrix}\right.\)
\( \Leftrightarrow \)\(\left\{\begin{matrix}U_{1}=3
& & \\ d=2
& &
\end{matrix}\right.\)
b) \( S=2U_{1}+2U_{2}+...+2U_{100}+U_{101}+U_{102}+...+U_{200}\)
\( \Leftrightarrow
S=U_{1}+U_{2}+...+U_{100}+U_{1}+U_{2}+...+U_{100}+U_{101}+U_{102}+...+U_{200}\)
\( \Leftrightarrow S=S_{100}+S_{200}\)
\(S_{100}=n.U_{1}+\frac{n(n-1)d}{2}=100.3+\frac{100.99.2}{2}=10200\)
\(S_{200}=200.3+\frac{200.199.2}{2}=40400\)
Vậy S=40400+10200=50600
