ข้อสอบคณิต O-NET - 2551

$$\begin{gathered} \mathrm{จากสูตร} \\ -\mathrm{ข้อมูลดิบ}\hspace{0.17em}\hspace{0.17em}\mathrm{ตำแหน่ง} Q_r = \frac{r}{4}·\left(N+1\right) \\ -\mathrm{ลำดับเลขคณิต}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{a}_n = a_1+\left(n-1\right)d \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{จะได้} \\ 92 = 2+\left(n-1\right)\left(3\right) \\ 90 = \left(n-1\right)\left(3\right) \\ 30 = n-1 \\ n = 31 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \mathrm{ตำแหน่ง} Q_3 = \frac{3}{4}\left(31+1\right) = 24 \\ \mathrm{ดังนั้น} a_{24} = 2+\left(24-1\right)\left(3\right) \\ = 2+69 \\ = 71 \end{gathered}$$