Bài tập nâng cao · Bài 8
Bài tập nâng cao Cấp số cộng · Bài 8
Xác định số hạng đầu, công sai và số hạng thứ n của các cấp số cộng sau, biết rằng:
a). $\left\{ \begin{array}{l} S_{12}=34 \\ S_{18}=45 \end{array} \right.$
b). $\left\{ \begin{array}{l} u_5=10 \\ S_{10}=5 \end{array} \right.$
c). $\frac{S_{20}}{5}=\frac{S_{10}}{3}=\frac{S_5}{2}$
d). $\left\{ \begin{array}{l} S_{20}=2S_{10} \\ S_{15}=3S_5 \end{array} \right.$
a). $\left\{ \begin{array}{l} S_{12}=34 \\ S_{18}=45 \end{array} \right.$
b). $\left\{ \begin{array}{l} u_5=10 \\ S_{10}=5 \end{array} \right.$
c). $\frac{S_{20}}{5}=\frac{S_{10}}{3}=\frac{S_5}{2}$
d). $\left\{ \begin{array}{l} S_{20}=2S_{10} \\ S_{15}=3S_5 \end{array} \right.$
Xem lời giải
Lời giải
a). $\left\{ \begin{array}{l} S_{12}=34 \\ S_{18}=45 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} \frac{12\left( 2u_1+11d \right)}{2}=34 \\ \frac{18\left( 2u_1+17d \right)}{2}=45 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} 6u_1+33d=17 \\ 2u_1+17d=5 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1=\frac{31}{9} \\ d=-\frac{1}{9} \end{array} \right.$
$u_n=u_1+\left( n-1 \right)d=\frac{32}{9}-\frac{1}{9}n$
b). $\left\{ \begin{array}{l} u_5=10 \\ S_{10}=5 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1+4d=10 \\ \frac{10\left( 2u_1+9d \right)}{2}=5 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1+4d=10 \\ 2u_1+9d=1 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1=86 \\ d=-19 \end{array} \right.$
$u_n=u_1+\left( n-1 \right)d=105-19n$
c). $\frac{S_{20}}{5}=\frac{S_{10}}{3}=\frac{S_5}{2}\Leftrightarrow \left\{ \begin{array}{l} \frac{S_{20}}{5}=\frac{S_{10}}{3} \\ \frac{S_{10}}{3}=\frac{S_5}{2} \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} \frac{20\left( 2u_1+19d \right)}{10}=\frac{10\left( 2u_1+9d \right)}{6} \\ \frac{10\left( 2u_1+9d \right)}{6}=\frac{5\left( 2u_1+4d \right)}{4} \end{array} \right.$
$\left\{ \begin{array}{l} 2u_1+69d=0 \\ 2u_1+24d=0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1=0 \\ d=0 \end{array} \right.\Rightarrow u_n=0$
d). $\left\{ \begin{array}{l} S_{20}=2S_{10} \\ S_{15}=3S_5 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} 20\left( 2u_1+19d \right)=20\left( 2u_1+9d \right) \\ 15\left( 2u_1+14d \right)=15\left( 2u_1+4d \right) \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} d=0 \\ u_1\in \mathbb{R} \end{array} \right.\Rightarrow u_n=u_1\in \mathbb{R}$
$u_n=u_1+\left( n-1 \right)d=\frac{32}{9}-\frac{1}{9}n$
b). $\left\{ \begin{array}{l} u_5=10 \\ S_{10}=5 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1+4d=10 \\ \frac{10\left( 2u_1+9d \right)}{2}=5 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1+4d=10 \\ 2u_1+9d=1 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1=86 \\ d=-19 \end{array} \right.$
$u_n=u_1+\left( n-1 \right)d=105-19n$
c). $\frac{S_{20}}{5}=\frac{S_{10}}{3}=\frac{S_5}{2}\Leftrightarrow \left\{ \begin{array}{l} \frac{S_{20}}{5}=\frac{S_{10}}{3} \\ \frac{S_{10}}{3}=\frac{S_5}{2} \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} \frac{20\left( 2u_1+19d \right)}{10}=\frac{10\left( 2u_1+9d \right)}{6} \\ \frac{10\left( 2u_1+9d \right)}{6}=\frac{5\left( 2u_1+4d \right)}{4} \end{array} \right.$
$\left\{ \begin{array}{l} 2u_1+69d=0 \\ 2u_1+24d=0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1=0 \\ d=0 \end{array} \right.\Rightarrow u_n=0$
d). $\left\{ \begin{array}{l} S_{20}=2S_{10} \\ S_{15}=3S_5 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} 20\left( 2u_1+19d \right)=20\left( 2u_1+9d \right) \\ 15\left( 2u_1+14d \right)=15\left( 2u_1+4d \right) \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} d=0 \\ u_1\in \mathbb{R} \end{array} \right.\Rightarrow u_n=u_1\in \mathbb{R}$