ข้อสอบ PAT 1 - มีนาคม 2556

$$\begin{gathered} \mathrm{จากโจทย์} {\mathrm{a}}_1 , {\mathrm{a}}_2 , {\mathrm{a}}_3 , {\mathrm{a}}_4 , {\mathrm{a}}_5 \mathrm{และ} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{b}}_{1 }, {\mathrm{b}}_2 , {\mathrm{b}}_3 , {\mathrm{b}}_{4 }, {\mathrm{b}}_{5 }, {\mathrm{b}}_6 \mathrm{เป็นลำดับเลขคณิต} \\ \mathrm{จะได้} {\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_1+\left(\mathrm{n}-1\right){\mathrm{d}}_{\mathrm{a}} \mathrm{และ} {\mathrm{b}}_{\mathrm{n}}={\mathrm{b}}_1+\left(\mathrm{n}-1\right){\mathrm{d}}_{\mathrm{b}} \\ \mathrm{จากโจทย์} {\mathrm{a}}_1={\mathrm{b}}_2 \\ \mathrm{จะได้} {\mathrm{a}}_1={\mathrm{b}}_1+{\mathrm{d}}_{\mathrm{b}}\to \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} {\mathrm{a}}_5={\mathrm{b}}_5 \\ \mathrm{จะได้} {\mathrm{a}}_1+4{\mathrm{d}}_{\mathrm{a}}={\mathrm{b}}_1+4{\mathrm{d}}_{\mathrm{b}}\to \boxed{2} \\ \mathrm{นำ} \boxed{2}-\boxed{1} \mathrm{จะได้} 4{\mathrm{d}}_{\mathrm{a}}=3{\mathrm{d}}_{\mathrm{b}} \\ \frac{{\mathrm{d}}_{\mathrm{b}}}{{\mathrm{d}}_{\mathrm{a}}}=\frac{4}{3} \end{gathered}$$​

$$\begin{gathered} \mathrm{จากโจทย์} \frac{\left({\mathrm{b}}_6-{\mathrm{b}}_4\right)+\left({\mathrm{b}}_6-{\mathrm{b}}_1\right)}{{\mathrm{a}}_4-{\mathrm{a}}_2}= \frac{\mathrm{x}}{\mathrm{y}}\hspace{0.17em} \\ \frac{2{\mathrm{b}}_6-{\mathrm{b}}_4-{\mathrm{b}}_1}{{\mathrm{a}}_4-{\mathrm{a}}_2}=\frac{\mathrm{x}}{\mathrm{y}} \\ \frac{2\left[{\mathrm{b}}_1+5{\mathrm{d}}_{\mathrm{b}}\right]-\left[{\mathrm{b}}_1+3{\mathrm{d}}_{\mathrm{b}}\right]-{\mathrm{b}}_1}{\left[{\mathrm{a}}_1+3{\mathrm{d}}_{\mathrm{a}}\right]-\left[{\mathrm{a}}_1+{\mathrm{d}}_{\mathrm{a}}\right]}=\frac{\mathrm{x}}{\mathrm{y}} \end{gathered}$$

$$\begin{gathered} \frac{7{\mathrm{d}}_{\mathrm{b}}}{2{\mathrm{d}}_{\mathrm{a}}}=\frac{\mathrm{x}}{\mathrm{y}} ; \mathrm{โดย} \frac{{\mathrm{d}}_{\mathrm{b}}}{{\mathrm{d}}_{\mathrm{a}}}=\frac{4}{3} \\ \frac{7}{2}\left(\frac{4}{3}\right)=\frac{\mathrm{x}}{\mathrm{y}} \\ \frac{\mathrm{x}}{\mathrm{y}}=\frac{14}{3} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{ห}.\mathrm{ร}.\mathrm{ม}. \mathrm{ของ} \mathrm{x} \mathrm{กับ} \mathrm{y} \mathrm{เท่ากับ} 1 \\ \mathrm{จะได้} \mathrm{x}=14 , \mathrm{y}=3 \\ \mathrm{ดังนั้น} {\mathrm{x}}^2+{\mathrm{y}}^2={14}^2+3^2 \\ =196+9 \\ =205 \end{gathered}$$