解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=95$$ 和公比：$$r=0.6842105263157895$$ 插入几何级数的公式：

$$a_1=95$$

$$a_2=a_1·r^{n-1}=95\cdot {0.6842105263157895}^{2-1}=95\cdot {0.6842105263157895}^1=95\cdot 0.6842105263157895=65$$

$$a_3=a_1·r^{n-1}=95\cdot {0.6842105263157895}^{3-1}=95\cdot {0.6842105263157895}^2=95\cdot 0.46814404432132967=44.473684210526315$$

$$a_4=a_1·r^{n-1}=95\cdot {0.6842105263157895}^{4-1}=95\cdot {0.6842105263157895}^3=95\cdot 0.32030908295669924=30.42936288088643$$

$$a_5=a_1·r^{n-1}=95\cdot {0.6842105263157895}^{5-1}=95\cdot {0.6842105263157895}^4=95\cdot 0.21915884623353107=20.820090392185453$$

$$a_6=a_1·r^{n-1}=95\cdot {0.6842105263157895}^{6-1}=95\cdot {0.6842105263157895}^5=95\cdot 0.14995078952820548=14.24532500517952$$

$$a_7=a_1·r^{n-1}=95\cdot {0.6842105263157895}^{7-1}=95\cdot {0.6842105263157895}^6=95\cdot 0.10259790862456164=9.746801319333356$$

$$a_8=a_1·r^{n-1}=95\cdot {0.6842105263157895}^{8-1}=95\cdot {0.6842105263157895}^7=95\cdot 0.0701985690589106=6.6688640605965075$$

$$a_9=a_1·r^{n-1}=95\cdot {0.6842105263157895}^{9-1}=95\cdot {0.6842105263157895}^8=95\cdot 0.04803059988241252=4.562906988829189$$

$$a_{10}=a_1·r^{n-1}=95\cdot {0.6842105263157895}^{10-1}=95\cdot {0.6842105263157895}^9=95\cdot 0.03286304202480857=3.121988992356814$$