ข้อสอบคณิต O-NET - 2554

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

$$\begin{gathered} \mathrm{จะได้} S_5 = \frac{5}{2}\left[2a_1+\left(5-1\right)d\right] = 90 \\ 2a_1+4d = 36 \dots \overline{)1} \\ \mathrm{จะได้} S_{10} = \frac{10}{2}\left[2a_1+\left(10-1\right)d\right] = 5 \\ 2a_1+9d = 1 \dots \overline{)2} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \overline{)1} - \overline{)2} ; 35 = -5d \\ d = -7 \\ \mathrm{แทนค่า} d=-7 \mathrm{ใน} \overline{)2} \\ \mathrm{จะได้} 1 = 2a_1+9\left(-7\right) \\ 64 = 2a_1 \\ a_1 = 32 \end{gathered}$$

$$\begin{gathered} \mathrm{จากสูตร} a_n = a_1+\left(n-1\right)d \\ a_{11} = 32+\left(11-1\right)\left(-7\right) \\ = 32+\left(-70\right) \\ = -38 \\ \mathrm{ดังนั้น} \mathrm{ตอบข้อ} 2 \end{gathered}$$