Bài tập nâng cao · Bài 6
Bài tập nâng cao Cấp số cộng · Bài 6
Tìm số hạng đầu và công sai của cấp số cộng, biết:
a). $\left\{ \begin{array}{l} u_7=27 \\ u_{15}=59 \end{array} \right.$
b). $\left\{ \begin{array}{l} u_9=5u_2 \\ u_{13}=2u_6+5 \end{array} \right.$
c). $\left\{ \begin{array}{l} u_2+u_4-u_6=-7 \\ u_8-u_7=2u_4 \end{array} \right.$
d). $\left\{ \begin{array}{l} u_3-u_7=-8 \\ u_2.u_7=75 \end{array} \right.$
e). $\left\{ \begin{array}{l} u_1^{2}+u_2^{2}+u_3^{2}=155 \\ S_3=21 \end{array} \right.$
a). $\left\{ \begin{array}{l} u_7=27 \\ u_{15}=59 \end{array} \right.$
b). $\left\{ \begin{array}{l} u_9=5u_2 \\ u_{13}=2u_6+5 \end{array} \right.$
c). $\left\{ \begin{array}{l} u_2+u_4-u_6=-7 \\ u_8-u_7=2u_4 \end{array} \right.$
d). $\left\{ \begin{array}{l} u_3-u_7=-8 \\ u_2.u_7=75 \end{array} \right.$
e). $\left\{ \begin{array}{l} u_1^{2}+u_2^{2}+u_3^{2}=155 \\ S_3=21 \end{array} \right.$
Xem lời giải
Lời giải
Gọi số hạng đầu là $u_1$ và công sai là $d$.
a).$\left\{ \begin{array}{l} u_7=27 \\ u_{15}=59 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1+6d=27 \\ u_1+14d=59 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1=3 \\ d=4 \end{array} \right.$
b). $\left\{ \begin{array}{l} u_9=5u_2 \\ u_{13}=2u_6+5 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1+8d=5\left( u_1+d \right) \\ u_1+12d=2\left( u_1+5d \right)+5 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} 4u_1-3d=0 \\ u_1-2d=-5 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1=3 \\ d=4 \end{array} \right.$
c) $\left\{ \begin{array}{l} u_2+u_4-u_6=-7 \\ u_8-u_7=2u_4 \end{array} \right. \left( 1 \right)$
$\left( 1 \right)\Leftrightarrow \left\{ \begin{array}{l} u_1+d+u_1+3d-\left( u_1+5d \right)=-7 \\ u_1+7d-\left( u_1+6d \right)=2\left( u_1+3d \right) \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1-d=-7 \\ 2u_1+5d=0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1=-5 \\ d=2. \end{array} \right.$
d) $\left\{ \begin{array}{l} u_3-u_7=-8 \\ u_2.u_7=75 \end{array} \right. \left( 1 \right)$
$\left( 1 \right)\Leftrightarrow \left\{ \begin{array}{l} u_1+2d-\left( u_1+6d \right)=-8 \\ \left( u_1+d \right)\left( u_1+6d \right)=75 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} -4d=-8 \\ \left( u_1+d \right)\left( u_1+6d \right)=75 \end{array} \right.\left\{ \begin{array}{l} d=2 \\ \left( u_1+2 \right)\left( u_1+12 \right)=75 \left( * \right) \end{array} \right.$
Giải $\left( * \right)\Leftrightarrow u_1^{2}+14u_1-51=0\Leftrightarrow \left\{ \begin{array}{l} u_1=3 \\ u_1=-17 \end{array} \right.$
Vậy $\left\{ \begin{array}{l} u_1=3 \\ d=2 \end{array} \right.$ hoặc $\left\{ \begin{array}{l} u_1=-17 \\ d=2. \end{array} \right.$
e). $\left\{ \begin{array}{l} u_1^{2}+u_2^{2}+u_3^{2}=155 \\ S_3=21 \end{array} \right.$
Ta có: $S_3=21\Leftrightarrow u_1+u_2+u_3=21\Leftrightarrow u_1+u_1+d+u_1+2d=21\Leftrightarrow d=7-u_1.$
Ta có: $u_1^{2}+u_2^{2}+u_3^{2}=155\Leftrightarrow u_1^{2}+\left( u_1+d \right)^{2}+\left( u_1+2d \right)^{2}=155$
$\Leftrightarrow u_1^{2}+\left( u_1+7-u_1 \right)^{2}+\left( u_1+14-2u_1 \right)^{2}=155\Leftrightarrow u_1^{2}+49+\left( 14-u_1 \right)^{2}=155$
$\Leftrightarrow 2u_1^{2}-28u_1+90=0\Leftrightarrow u_1=9 \vee u_1=5$
Với $u_1=9\Rightarrow d=-2$. Với $u_1=5\Rightarrow d=2$
a).$\left\{ \begin{array}{l} u_7=27 \\ u_{15}=59 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1+6d=27 \\ u_1+14d=59 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1=3 \\ d=4 \end{array} \right.$
b). $\left\{ \begin{array}{l} u_9=5u_2 \\ u_{13}=2u_6+5 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1+8d=5\left( u_1+d \right) \\ u_1+12d=2\left( u_1+5d \right)+5 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} 4u_1-3d=0 \\ u_1-2d=-5 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1=3 \\ d=4 \end{array} \right.$
c) $\left\{ \begin{array}{l} u_2+u_4-u_6=-7 \\ u_8-u_7=2u_4 \end{array} \right. \left( 1 \right)$
$\left( 1 \right)\Leftrightarrow \left\{ \begin{array}{l} u_1+d+u_1+3d-\left( u_1+5d \right)=-7 \\ u_1+7d-\left( u_1+6d \right)=2\left( u_1+3d \right) \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1-d=-7 \\ 2u_1+5d=0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} u_1=-5 \\ d=2. \end{array} \right.$
d) $\left\{ \begin{array}{l} u_3-u_7=-8 \\ u_2.u_7=75 \end{array} \right. \left( 1 \right)$
$\left( 1 \right)\Leftrightarrow \left\{ \begin{array}{l} u_1+2d-\left( u_1+6d \right)=-8 \\ \left( u_1+d \right)\left( u_1+6d \right)=75 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} -4d=-8 \\ \left( u_1+d \right)\left( u_1+6d \right)=75 \end{array} \right.\left\{ \begin{array}{l} d=2 \\ \left( u_1+2 \right)\left( u_1+12 \right)=75 \left( * \right) \end{array} \right.$
Giải $\left( * \right)\Leftrightarrow u_1^{2}+14u_1-51=0\Leftrightarrow \left\{ \begin{array}{l} u_1=3 \\ u_1=-17 \end{array} \right.$
Vậy $\left\{ \begin{array}{l} u_1=3 \\ d=2 \end{array} \right.$ hoặc $\left\{ \begin{array}{l} u_1=-17 \\ d=2. \end{array} \right.$
e). $\left\{ \begin{array}{l} u_1^{2}+u_2^{2}+u_3^{2}=155 \\ S_3=21 \end{array} \right.$
Ta có: $S_3=21\Leftrightarrow u_1+u_2+u_3=21\Leftrightarrow u_1+u_1+d+u_1+2d=21\Leftrightarrow d=7-u_1.$
Ta có: $u_1^{2}+u_2^{2}+u_3^{2}=155\Leftrightarrow u_1^{2}+\left( u_1+d \right)^{2}+\left( u_1+2d \right)^{2}=155$
$\Leftrightarrow u_1^{2}+\left( u_1+7-u_1 \right)^{2}+\left( u_1+14-2u_1 \right)^{2}=155\Leftrightarrow u_1^{2}+49+\left( 14-u_1 \right)^{2}=155$
$\Leftrightarrow 2u_1^{2}-28u_1+90=0\Leftrightarrow u_1=9 \vee u_1=5$
Với $u_1=9\Rightarrow d=-2$. Với $u_1=5\Rightarrow d=2$