ข้อสอบ PAT 1 - มีนาคม 2555

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} {\mathrm{a}}_1+{\mathrm{a}}_2=20 \\ \mathrm{จะได้} {\mathrm{a}}_1+{\mathrm{a}}_1\mathrm{r}=20 \\ {\mathrm{a}}_1\left(1+\mathrm{r}\right)=20\to \boxed{1} \\ \mathrm{จากโจทย์}\hspace{0.17em} {\mathrm{a}}_1+{\mathrm{a}}_2+{\mathrm{a}}_3+{\mathrm{a}}_4=65 \\ \mathrm{จะได้} {\mathrm{a}}_3+{\mathrm{a}}_4=45 \\ {\mathrm{a}}_1{\mathrm{r}}^2+{\mathrm{a}}_1{\mathrm{r}}^3=45 \\ {\mathrm{a}}_1{\mathrm{r}}^2\left(1+\mathrm{r}\right)=45\to \boxed{2} \end{gathered}$$

$$\begin{gathered} \mathrm{นำ} \frac{\boxed{2}}{\boxed{1}} {\mathrm{r}}^2=\frac{45}{20} \\ {\mathrm{r}}^2=\frac{9}{4} \\ \mathrm{r}=\pm \frac{3}{2} \\ \mathrm{จากโจทย์} \mathrm{อัตราส่วนร่วมเป็นจำนวนจริงบวก} \\ \mathrm{จะได้}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}r=\frac{3}{2} \\ -\mathbf{แทน}\mathbf{ }\mathbf{ }\mathbf{r}\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}} \mathrm{ใน} \boxed{1} \\ \mathrm{จะได้} {\mathrm{a}}_1\left(1+\frac{3}{2}\right)=20 \\ {\mathrm{a}}_1=8 \end{gathered}$$

$$\begin{gathered} \mathrm{จากสูตร} {\mathrm{S}}_{\mathrm{n}}=\frac{{\mathrm{a}}_1\left({\mathrm{r}}^{\mathrm{n}}-1\right)}{\mathrm{r}-1} \mathrm{โดย} \left|\mathrm{r}\right|>1 \\ {\mathrm{S}}_{\mathrm{n}}=\frac{8\left({\left({\displaystyle \frac{3}{2}}\right)}^6-1\right)}{{\displaystyle \frac{3}{2}}-1} \\ =16\left(\frac{729}{64}-1\right) \\ =\frac{729}{4}-16 \\ =182.25-16 \\ =166.25 \end{gathered}$$