ข้อสอบ PAT 1 - มีนาคม 2558

$$\begin{gathered} \mathrm{จากโจทย์} \frac{{\mathrm{a}}_1+{\mathrm{a}}_2+...+{\mathrm{a}}_{\mathrm{n}}}{{\mathrm{b}}_1+{\mathrm{b}}_2+...+{\mathrm{b}}_{\mathrm{n}}}=\frac{\mathrm{n}+1}{2\mathrm{n}-1} \\ \mathrm{จากสูตร} \mathrm{อนุกรมเลขคณิต} {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2{\mathrm{a}}_1+\left(\mathrm{n}-1\right)\mathrm{d}\right] \\ \mathrm{จะได้} \frac{{\displaystyle \frac{\mathrm{n}}{2}}\left[2{\mathrm{a}}_1+\left(\mathrm{n}-1\right){\mathrm{d}}_{\mathrm{a}}\right]}{{\displaystyle \frac{\mathrm{n}}{2}}\left[2{\mathrm{b}}_1+\left(\mathrm{n}-1\right){\mathrm{d}}_{\mathrm{b}}\right]}=\frac{\mathrm{n}+1}{2\mathrm{n}-1} \end{gathered}$$

$$\begin{gathered} \mathrm{เมื่อ} \mathrm{n}=1 \frac{2{\mathrm{a}}_1+\left(1-1\right){\mathrm{d}}_{\mathrm{a}}}{2{\mathrm{b}}_1+\left(1-1\right){\mathrm{d}}_{\mathrm{b}}}=\frac{1+1}{2\left(1\right)-1} \\ \frac{2{\mathrm{a}}_1}{2{\mathrm{b}}_1}=2 \\ {\mathrm{a}}_1=2{\mathrm{b}}_1\to \boxed{1} \\ \mathrm{เมื่อ} \mathrm{n}=2 \frac{2{\mathrm{a}}_1+\left(2-1\right){\mathrm{d}}_{\mathrm{a}}}{2{\mathrm{b}}_1+\left(2-1\right){\mathrm{d}}_{\mathrm{b}}}=\frac{1+1}{2\left(2\right)-1} \\ \frac{2{\mathrm{a}}_1+{\mathrm{d}}_{\mathrm{a}}}{2{\mathrm{b}}_1+{\mathrm{d}}_{\mathrm{b}}}=1 \\ 2{\mathrm{a}}_1+{\mathrm{d}}_{\mathrm{a}}=2{\mathrm{b}}_1+{\mathrm{d}}_{\mathrm{b}} \end{gathered}$$

$$\begin{gathered} \mathrm{แทน} {\mathrm{a}}_1=2{\mathrm{b}}_1 \\ \mathrm{จะได้} 2\left(2{\mathrm{b}}_1\right)+{\mathrm{d}}_{\mathrm{a}}=2{\mathrm{b}}_1+{\mathrm{d}}_{\mathrm{b}} \\ 2{\mathrm{b}}_1+{\mathrm{d}}_{\mathrm{a}}={\mathrm{d}}_{\mathrm{b}}\to \boxed{2} \\ \mathrm{เมื่อ} \mathrm{n}=3 \frac{2{\mathrm{a}}_1+\left(3-1\right){\mathrm{d}}_{\mathrm{a}}}{2{\mathrm{b}}_1+\left(3-1\right){\mathrm{d}}_{\mathrm{b}}}=\frac{3+1}{2\left(3\right)-1} \\ \frac{2{\mathrm{a}}_1+2{\mathrm{d}}_{\mathrm{a}}}{2{\mathrm{b}}_1+2{\mathrm{d}}_{\mathrm{b}}}=\frac{4}{5} \\ \frac{2\left({\mathrm{a}}_1+{\mathrm{d}}_{\mathrm{a}}\right)}{2\left({\mathrm{b}}_1+{\mathrm{d}}_{\mathrm{b}}\right)}=\frac{4}{5} ; \mathrm{จาก} \boxed{1} \mathrm{แทน} {\mathrm{a}}_1=2{\mathrm{b}}_1 \end{gathered}$$

$$\begin{gathered} \frac{2{\mathrm{a}}_1+{\mathrm{d}}_{\mathrm{a}}}{{\mathrm{b}}_1+{\mathrm{d}}_{\mathrm{b}}}=\frac{4}{5} ; \mathrm{จาก} \boxed{2} \mathrm{แทน} 2{\mathrm{b}}_1+{\mathrm{d}}_{\mathrm{a}}={\mathrm{d}}_{\mathrm{b}} \\ \frac{{\mathrm{d}}_{\mathrm{b}}}{{\mathrm{b}}_1+{\mathrm{d}}_{\mathrm{b}}}=\frac{4}{5} \\ 5{\mathrm{d}}_{\mathrm{b}}=4\left({\mathrm{b}}_1+{\mathrm{d}}_{\mathrm{b}}\right) \\ {\mathrm{d}}_{\mathrm{b}}=4{\mathrm{b}}_1\to \boxed{3} \\ \mathrm{จาก} \boxed{2} \mathrm{แทน} {\mathrm{d}}_{\mathrm{b}}=2{\mathrm{b}}_1+{\mathrm{d}}_{\mathrm{a}} \\ \mathrm{จะได้} 2{\mathrm{b}}_1+{\mathrm{d}}_{\mathrm{a}}=4{\mathrm{b}}_1 \\ {\mathrm{d}}_{\mathrm{a}}=2{\mathrm{b}}_1\to \boxed{4} \end{gathered}$$

$$\begin{gathered} \mathrm{จากสูตร} \mathrm{อนุกรมเลขคณิต} {\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_1+\left(\mathrm{n}-1\right)\mathrm{d} \\ \mathrm{ดังนั้น} \frac{2{\mathrm{b}}_{100}}{{\mathrm{a}}_{100}}=\frac{2\left[{\mathrm{b}}_1+99{\mathrm{d}}_{\mathrm{b}}\right]}{{\mathrm{a}}_1+99{\mathrm{d}}_{\mathrm{a}}} \\ \mathrm{จาก} \boxed{1},\boxed{3},\boxed{4} \mathrm{แทนค่า} \\ \mathrm{จะได้} \frac{2{\mathrm{b}}_{100}}{{\mathrm{a}}_{100}}=\frac{2\left[{\mathrm{b}}_1+99\left(4{\mathrm{b}}_1\right)\right]}{2{\mathrm{b}}_1+99\left(2{\mathrm{b}}_1\right)} \\ =\frac{\left(2\right)\left(397{\mathrm{b}}_1\right)}{200{\mathrm{b}}_1} \\ =\frac{397}{100} \\ =3.97 \end{gathered}$$