ข้อสอบคณิต O-NET - 2556

$$\begin{gathered} \mathrm{จากสูตรลำดับเลขคณิต} \\ a_n = a\left(n-1\right)d \\ \mathrm{จะได้} a_5 = a_1+4d \dots \overline{)1} \\ a_{10} = a_1+9d \dots \overline{)2} \end{gathered}$$

$$\begin{gathered} \overline{)2} -\overline{)1} ; 5d = 15 \\ d = 3 \\ \mathrm{แทน} d=3 \mathrm{ใน} \overline{)1} \\ a_1+4\left(3\right) = 14 \\ a_1 = 2 \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} a_{99} = a_1+98d \\ = 2+98\left(3\right) \\ = 296 \\ \mathrm{ตอบ} \mathrm{ข้อ} 3 \end{gathered}$$