ข้อสอบคณิต O-NET - 2563

$$\begin{gathered} \mathrm{จากสูตร} \mathrm{พจน์ทั่วไปลำดับเลขคณิต} \\ a_n = a_1+\left(n-1\right)d \\ \mathrm{จากสูตร} \mathrm{อนุกรมเลขคณิต} \\ S_n = \frac{n}{2}\left(a_1+a_n\right) \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} 2, a, 10, \dots , b \\ \mathrm{จะได้} a = a_2 \\ a = a_1+d \dots \overline{)1} \\ 10 = a_3 \\ 10 = a_1+2d \\ 10 = 2+2d \\ d = 4 \mathrm{แทนใน} \overline{)1} \end{gathered}$$

$$\begin{gathered} a = 6 \\ \mathrm{จะได้} S_n = \frac{n}{2}\left(a_1+a_n\right) \dots \overline{)2} \end{gathered}$$

$$\begin{gathered} \mathrm{หาจำนวนตัว} \left(n\right) \mathrm{ก่อน} \\ \mathrm{จะได้} n = \frac{a_n-a_1}{d}+1 \\ = \frac{b-2}{4}+1 \\ n = \frac{b+2 }{4} \mathrm{แทนใน} \overline{)2} \end{gathered}$$

$$\begin{gathered} 288 = \frac{\left({\displaystyle \frac{b+2}{4}}\right)}{2}\left(2+b\right) \\ {\left(b+2\right)}^2 = 288\times 8 \\ b+2 = \sqrt{288\times 8} \\ b = 46 \\ \mathrm{ดังนั้น} a+b = 6+46 = 52 \to \mathrm{ตอบข้อ} 3 \end{gathered}$$