解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=75$$ 和公比：$$r=1.2$$ 插入几何级数的公式：

$$a_1=75$$

$$a_2=a_1·r^{n-1}=75\cdot {1.2}^{2-1}=75\cdot {1.2}^1=75\cdot 1.2=90$$

$$a_3=a_1·r^{n-1}=75\cdot {1.2}^{3-1}=75\cdot {1.2}^2=75\cdot 1.44=108$$

$$a_4=a_1·r^{n-1}=75\cdot {1.2}^{4-1}=75\cdot {1.2}^3=75\cdot 1.7279999999999998=129.6$$

$$a_5=a_1·r^{n-1}=75\cdot {1.2}^{5-1}=75\cdot {1.2}^4=75\cdot 2.0736=155.51999999999998$$

$$a_6=a_1·r^{n-1}=75\cdot {1.2}^{6-1}=75\cdot {1.2}^5=75\cdot 2.4883199999999994=186.62399999999997$$

$$a_7=a_1·r^{n-1}=75\cdot {1.2}^{7-1}=75\cdot {1.2}^6=75\cdot 2.9859839999999993=223.94879999999995$$

$$a_8=a_1·r^{n-1}=75\cdot {1.2}^{8-1}=75\cdot {1.2}^7=75\cdot 3.583180799999999=268.73855999999995$$

$$a_9=a_1·r^{n-1}=75\cdot {1.2}^{9-1}=75\cdot {1.2}^8=75\cdot 4.2998169599999985=322.4862719999999$$

$$a_{10}=a_1·r^{n-1}=75\cdot {1.2}^{10-1}=75\cdot {1.2}^9=75\cdot 5.1597803519999985=386.9835263999999$$