Tìm số hạng đầu và công sai của cấp số cộng (un) biết:

c) $\left \{ {{u_2.u_5=52} \atop {u_2+u_3+u_4+u_5=34}} \right.$ 

Ta có:

`u_n=u_1+(n-1)d`

`=>u_2.u_5=52`

`<=>(u_1+d)(u_1+4d)=52`

`<=>u_1^{2}+5d.u_1+4d^{2}=52` (1)

và `u_2+u_3+u_4+u_5=34`

`<=>(u_1+d)+(u_1+2d)+(u_1+3d)+(u_1+4d)=34`

`<=>4u_1+10d=34`

`=>2u_1+5d=17`

`=>u_1=\frac{17-5d}{2}` thay vào (1) ta có:

`(\frac{17-5d}{2})^{2}+5d(\frac{17-5d}{2})+4d^{2}=52`

`=>(17-5d)^{2}+10d(17-5d)+16d^{2}=208`

`=>-9d^{2}=-81`

`=>d^{2}=9`

`=>d=±3`

`<=>` \(\left[ \begin{array}{l}d=3\\d=-3\end{array} \right.\) 

`=>`\(\left[ \begin{array}{l}u_1=1\\u_1=16\end{array} \right.\) 

Vậy \(\left[ \begin{array}{l}\left \{ {{u_1=1} \atop {d=3}} \right. \\\left \{ {{u_1=16} \atop {d=-3}} \right. \end{array} \right.\) 

d) $\left \{ {{u_7+u_{15}=60} \atop {u_4^{2}+u_{12}^{2}=1170}} \right.$ 

Ta có: `u_n=u_1+(n-1)d`

`=>u_7+u_{15}=60`

`<=>(u_1+6d)+(u_1+14d)=60`

`=>2u_1+20d=60`

`=>u_1+10d=30`

`=>u_1=30-10d`

và `u_4^{2}+u_{12}^{2}=1170`

`<=>(u_1+3d)^{2}+(u_1+11d)^{2}=1170`

`<=>(30-10d+3d)^{2}+(30-10d+11d)^{2}=1170`

`<=>(30-7d)^{2}+(30+d)^{2}=1170`

`=>`\(\left[ \begin{array}{l}d=3\\d=\frac{21}{5}\end{array} \right.\) 

`=>`\(\left[ \begin{array}{l}u_1=0\\u_1=-12\end{array} \right.\) 

Vậy \(\left[ \begin{array}{l}\left \{ {{u_1=0} \atop {d=3}} \right. \\\left \{ {{u_1=-12} \atop {d=\frac{21}{5}}} \right. \end{array} \right.\)