ข้อสอบคณิต O-NET - 2559

$$\begin{gathered} \mathrm{จากสูตรลำดับเลขคณิต} \\ a_n = a_1+\left(n-1\right)d \\ \mathrm{โจทย์ให้} a_4 = 5a_1 \\ a_{10} = 39 \\ \mathrm{แทน} n=4 : a_4 = a_1+3d \\ 5a_1 = a_1+3d \\ 4a_1 = 3d \dots \overline{)1} \end{gathered}$$

$$\begin{gathered} \mathrm{แทน} n=10 : a_{10} = a_1+9d \\ 39 = a_1+9d \dots \overline{)2} \\ \mathrm{เอา} \overline{)1} \times 3 \mathrm{แล้วแทนใน} \overline{)2} \\ 39 = a_1+12a_1 \\ 39 = 13a_1 \\ a_1 = 3 \\ \mathrm{ดังนั้น} \mathrm{ตอบข้อ} 3 \end{gathered}$$