Bài tập nâng cao · Bài 9
Bài tập nâng cao Cấp số cộng · Bài 9
Cho cấp số cộng: $u_1;u_2;u_3;\ldots$ có công sai d.
1). Biết $u_2+u_{22}=40.$ Tính $S_{23}$
2). Biết $u_1+u_4+u_7+u_{10}+u_{13}+u_{16}=147.$ Tính $u_6+u_{11} \vee u_1+u_6+u_{11}+u_{16}$
4). Biết $u_4+u_8+u_{12}+u_{16}=224.$ Tính: $S_{19}$
5). Biết $u_{23}+u_{57}=29$. Tính: $u_{10}+u_{70}+u_{157}+3u_1$
1). Biết $u_2+u_{22}=40.$ Tính $S_{23}$
2). Biết $u_1+u_4+u_7+u_{10}+u_{13}+u_{16}=147.$ Tính $u_6+u_{11} \vee u_1+u_6+u_{11}+u_{16}$
4). Biết $u_4+u_8+u_{12}+u_{16}=224.$ Tính: $S_{19}$
5). Biết $u_{23}+u_{57}=29$. Tính: $u_{10}+u_{70}+u_{157}+3u_1$
Xem lời giải
Lời giải
1). Biết $u_2+u_{22}=40.$ Tính $S_{23}$
Ta có: $u_2+u_{22}=40\Leftrightarrow u_1+d+u_1+21d=40\Leftrightarrow 2u_1+22d=40$
Mà $S_{23}=\frac{23}{2}\left( 2u_1+22d \right)=\frac{23}{2}.40=460.$
2). Biết $u_1+u_4+u_7+u_{10}+u_{13}+u_{16}=147.$ Tính $u_6+u_{11} \vee u_1+u_6+u_{11}+u_{16}$
Có: $u_1+u_4+u_7+u_{10}+u_{13}+u_{16}=147.$
$\Leftrightarrow u_1+u_1+3d+u_1+6d+u_1+9d+u_1+12d+u_1+15d=147.$
$\Leftrightarrow 6u_1+45d=147\Leftrightarrow 2u_1+15d=49.$
Ta có: $u_6+u_{11}=u_1+5d+u_1+10d=2u_1+15d=49.$
Ta có: $u_1+u_6+u_{11}+u_{16}=u_1+u_1+5d+u_1+10d+u_1+15d$
$=4u_1+30d=2\left( 2u_1+15d \right)=2.49=98.$
4). Biết $u_4+u_8+u_{12}+u_{16}=224.$ Tính: $S_{19}$
Có: $u_4+u_8+u_{12}+u_{16}=224$
$\Leftrightarrow u_1+3d+u_1+7d+u_1+11d+u_1+15d=224\Leftrightarrow 4u_1+36d=224\Leftrightarrow u_1+9d=56$
Ta có: $S_{19}=\frac{19}{2}\left( 2u_1+18d \right)=19\left( u_1+9d \right)=19.56=1064.$
5). Biết $u_{23}+u_{57}=29$. Tính: $u_{10}+u_{70}+u_{157}+3u_1$
Ta có: $u_{23}+u_{57}=29\Leftrightarrow u_1+22d+u_1+56d=29\Leftrightarrow 2u_1+78d=29.$
Ta có: $3u_1+u_{10}+u_{70}+u_{157}=3u_1+u_1+9d+u_1+69d+u_1+156d$
$=6u_1+234d=3\left( 2u_1+78d \right)=3.29=87$
Ta có: $u_2+u_{22}=40\Leftrightarrow u_1+d+u_1+21d=40\Leftrightarrow 2u_1+22d=40$
Mà $S_{23}=\frac{23}{2}\left( 2u_1+22d \right)=\frac{23}{2}.40=460.$
2). Biết $u_1+u_4+u_7+u_{10}+u_{13}+u_{16}=147.$ Tính $u_6+u_{11} \vee u_1+u_6+u_{11}+u_{16}$
Có: $u_1+u_4+u_7+u_{10}+u_{13}+u_{16}=147.$
$\Leftrightarrow u_1+u_1+3d+u_1+6d+u_1+9d+u_1+12d+u_1+15d=147.$
$\Leftrightarrow 6u_1+45d=147\Leftrightarrow 2u_1+15d=49.$
Ta có: $u_6+u_{11}=u_1+5d+u_1+10d=2u_1+15d=49.$
Ta có: $u_1+u_6+u_{11}+u_{16}=u_1+u_1+5d+u_1+10d+u_1+15d$
$=4u_1+30d=2\left( 2u_1+15d \right)=2.49=98.$
4). Biết $u_4+u_8+u_{12}+u_{16}=224.$ Tính: $S_{19}$
Có: $u_4+u_8+u_{12}+u_{16}=224$
$\Leftrightarrow u_1+3d+u_1+7d+u_1+11d+u_1+15d=224\Leftrightarrow 4u_1+36d=224\Leftrightarrow u_1+9d=56$
Ta có: $S_{19}=\frac{19}{2}\left( 2u_1+18d \right)=19\left( u_1+9d \right)=19.56=1064.$
5). Biết $u_{23}+u_{57}=29$. Tính: $u_{10}+u_{70}+u_{157}+3u_1$
Ta có: $u_{23}+u_{57}=29\Leftrightarrow u_1+22d+u_1+56d=29\Leftrightarrow 2u_1+78d=29.$
Ta có: $3u_1+u_{10}+u_{70}+u_{157}=3u_1+u_1+9d+u_1+69d+u_1+156d$
$=6u_1+234d=3\left( 2u_1+78d \right)=3.29=87$