ข้อสอบคณิต O-NET - 2558

$$\begin{gathered} \mathrm{จากสูตรลำดับเลขคณิต} \\ {a}_n = a_1+\left(n-1\right)d \\ {S}_n = \frac{n}{2}\left(a_1+a_n\right) \\ = \frac{n}{2}\left[2a_1+\left(n-1\right)d\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} S_9 = 261 , a_9 = 61 \\ \to \mathrm{รู้พจน์สุดท้าย} \mathrm{ใช้สูตร} S_n = \frac{n}{2}\left(a_1+a_n\right) \\ S_9 = \frac{9}{2}\left(a_1+a_9\right) \\ 261 = \frac{9}{2}\left(a_1+61\right) \\ 58 = a_1+61 \\ a_1 = -3 \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} a_9 = 61 , a_1 = -3 \\ \mathrm{จะหา} d \mathrm{จาก} a_n = a_1+\left(n-1\right)d \\ a_9 = a_1+\left(9-1\right)d \\ 61 = -3+8d \\ 64 = 8d \\ d = 8 \end{gathered}$$

$$\begin{gathered} \mathrm{โจทย์ถาม} S_4 \to \mathrm{แทน} n=4 , a_1 = -3 , d=8 \\ S_4 = \frac{4}{2}\left[2\left(-3\right)+\left(4-1\right)8\right] \\ = 2\left(-6+24\right) \\ = 36 \to \mathrm{ตอบข้อ} 4 \end{gathered}$$