ข้อสอบคณิต 2 - มีนาคม 2562

$$\begin{gathered} \mathrm{จากสูตรอนุกรมเลขคณิต} \\ S_n = \frac{n}{2}\left[2a_1+\left(n-1\right)d\right] \end{gathered}$$

$$\begin{gathered} \mathrm{แทน} n=10 \mathrm{จะได้} \\ S_{10} = \frac{10}{2}\left[2a_1+\left(10-1\right)d\right] \\ 55 = 10a_1+45d \\ 11 = 2a_1+9d \dots \overline{)1} \end{gathered}$$

$$\begin{gathered} \mathrm{แทน} n=11 \mathrm{จะได้} \\ S_{11} = \frac{11}{2}\left[2a_1+\left(11-1\right)d\right] \\ 77 = 11a_1+55d \\ 7 = a_1+5d \dots \overline{)2} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} 2\overline{)2} - \overline{)1} : 2\left(7\right)-11 = 2\left(a_1+5d\right)-\left(2a_1+9d\right) \\ d = 3 \mathrm{แทนใน} \overline{)2} \\ 7 = a_1+5\left(3\right) \\ a_1 = -8 \end{gathered}$$

$$\begin{gathered} \mathrm{แทน} n=9 \mathrm{จะได้} \\ S_9 = \frac{9}{2}\left[2\left(-8\right)+\left(9-1\right)\left(3\right)\right] = 36 \\ \mathrm{ดังนั้น} \mathrm{ตอบ} 3 \end{gathered}$$