ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2563 (วิชาสามัญเก่า)

$$\begin{gathered} \mathrm{จากสูตร} \mathrm{อนุกรมเรขาคณิต} \\ S_n = \frac{a_1\left(1-r^n\right)}{1-r} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} a_1+a_2+\dots +a_5 = \frac{a_1\left(1-r^5\right)}{1-r} \dots \overline{)1} = 4 \\ a_1+a_2+\dots +a_{15} = \frac{a_1\left(1-r^{15}\right)}{1-r} \\ 4+3 = \frac{a_1\left(1-r^{15}\right)}{1-r} \\ 7 = \frac{a_1\left(1-r^{15}\right)}{1-r} \dots \overline{)2} \end{gathered}$$

$$\begin{gathered} \frac{\overline{)2}}{\overline{)1}} ; \frac{7}{4} = \frac{a_1\left(1-r^{15}\right)}{1-r}·\frac{1-r}{a_1\left(1-r^5\right)} \\ \frac{7}{4} = \frac{1-r^{15}}{1-r^5} \\ \frac{7}{4} = \frac{\left(1-r^5\right)\left(1+r^5+r^{10}\right)}{\left(1-r^5\right)} \end{gathered}$$

$$\begin{gathered} 7 = 4+4r^5+4r^{10} \\ 4r^{10}+4r^5-3 = 0 \\ \left(2r^5-1\right)\left(2r^5+3\right) = 0 \\ r^5 = \frac{1}{2} , -\frac{3}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{เนื่องจาก} \left|r^5\right| < 1 \\ -1 < r^5 < 1 \\ -\frac{3}{2} \mathrm{ใช้ต่อไม่ได้} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} S_{\infty } = \frac{a_1}{1-r} \\ \mathrm{แทนค่า} r^5 \mathrm{ใน} \overline{)1} = \frac{a_1\left(1-{\displaystyle \frac{1}{2}}\right)}{1-r} = 4 \\ \mathrm{ดังนั้น} \sum _{n=1}^{\infty }{a}_n \frac{a_1}{1-r} = 8 \\ \mathrm{ตอบ} 1 \end{gathered}$$