解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=125$$ 和公比：$$r=1.2$$ 插入几何级数的公式：

$$a_1=125$$

$$a_2=a_1·r^{n-1}=125\cdot {1.2}^{2-1}=125\cdot {1.2}^1=125\cdot 1.2=150$$

$$a_3=a_1·r^{n-1}=125\cdot {1.2}^{3-1}=125\cdot {1.2}^2=125\cdot 1.44=180$$

$$a_4=a_1·r^{n-1}=125\cdot {1.2}^{4-1}=125\cdot {1.2}^3=125\cdot 1.7279999999999998=215.99999999999997$$

$$a_5=a_1·r^{n-1}=125\cdot {1.2}^{5-1}=125\cdot {1.2}^4=125\cdot 2.0736=259.2$$

$$a_6=a_1·r^{n-1}=125\cdot {1.2}^{6-1}=125\cdot {1.2}^5=125\cdot 2.4883199999999994=311.0399999999999$$

$$a_7=a_1·r^{n-1}=125\cdot {1.2}^{7-1}=125\cdot {1.2}^6=125\cdot 2.9859839999999993=373.24799999999993$$

$$a_8=a_1·r^{n-1}=125\cdot {1.2}^{8-1}=125\cdot {1.2}^7=125\cdot 3.583180799999999=447.8975999999999$$

$$a_9=a_1·r^{n-1}=125\cdot {1.2}^{9-1}=125\cdot {1.2}^8=125\cdot 4.2998169599999985=537.4771199999998$$

$$a_{10}=a_1·r^{n-1}=125\cdot {1.2}^{10-1}=125\cdot {1.2}^9=125\cdot 5.1597803519999985=644.9725439999999$$