ข้อสอบคณิต O-NET - 2560

$$\begin{gathered} \mathrm{จากโจทย์} S_n \mathrm{คือ} \mathrm{ผลบวก} n \mathrm{พจน์แรก} \\ \mathrm{นั่นคือ} S_1 = a_1 \\ S_2 = a_1+a_2 \\ S_3 = a_1+a_2+a_3 \\ ⋮ \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} S_n = n^2-4n \mathrm{แทน} n=1 \\ \mathrm{จะได้} S_1 = {\left(1\right)}^2-4\left(1\right) = -3 \\ \mathrm{ดังนั้น} a_1 = -3 \dots \overline{)1} \end{gathered}$$

$$\begin{gathered} \mathrm{แทน} n=2 \\ \mathrm{จะได้} S_2 = {\left(2\right)}^2-4\left(2\right) = -4 \\ \mathrm{ดังนั้น} a_1+a_2 = -4 \dots \overline{)2} \end{gathered}$$

$$\begin{gathered} \overline{)2} - \overline{)1} : a_2 = -4-\left(-3\right) = -1 \\ d = a_2-a_1 = \left(-1\right)-\left(-3\right) = 2 \\ \mathrm{ดังนั้น} d+a_1{a}_2 = 2+\left(-3\right)\left(-1\right) = 5 \end{gathered}$$