ข้อสอบคณิต 2 - มีนาคม 2562

$$\begin{gathered} \mathrm{จากสูตรอนุกรมเรขาคณิต} \\ S_n = \frac{a_1\left(1-r^n\right)}{1-r} \\ \mathrm{จะได้} a_1+a_2+\dots +a_{20} = S_{20} \\ = \frac{a_1\left(1-r^{20}\right)}{1-r} = 13 \dots \overline{)1} \end{gathered}$$

$$\begin{gathered} \mathrm{และจาก} a_1-a_2+a_3-a_4+\dots +a_{19}-a_{20} \\ = a_1+\left(-a_2\right)+a_3+\left(-a_4\right)+\dots +a_{19}+\left(-a_{20}\right) \\ \mathrm{จะได้} a_1+\left(-a_2\right)+a_3+\left(-a_4\right)+\dots +a_{19}+\left(-a_{20}\right) \\ = \frac{a_1\left[1-{\left(-r\right)}^{20}\right]}{1-\left(-r\right)} = \frac{a_1\left(1-r^{20}\right)}{1+r} = 17 \dots \overline{)2} \end{gathered}$$

$$\begin{gathered} \frac{\overline{)1}}{\overline{)2}} ; \frac{a_1\left(1-r^{20}\right)}{1-r}÷\frac{a_1\left(1-r^{20}\right)}{1+r} = \frac{13}{17} \\ \frac{\cancel{a_1\left(1-r^{20}\right)}}{1-r}\times \frac{1+r}{\cancel{a_1\left(1-r^{20}\right)}} = \frac{13}{17} \end{gathered}$$

$$\begin{gathered} \frac{1+r}{1-r} = \frac{13}{17} \\ 17+17r = 13-13r \\ 30r = -4 \\ r = -\frac{4}{30} = -\frac{2}{15} \\ \mathrm{ดังนั้น} \mathrm{ตอบ} 1 \end{gathered}$$