解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=910$$ 和公比：$$r=0.6153846153846154$$ 插入几何级数的公式：

$$a_1=910$$

$$a_2=a_1·r^{n-1}=910\cdot {0.6153846153846154}^{2-1}=910\cdot {0.6153846153846154}^1=910\cdot 0.6153846153846154=560$$

$$a_3=a_1·r^{n-1}=910\cdot {0.6153846153846154}^{3-1}=910\cdot {0.6153846153846154}^2=910\cdot 0.37869822485207105=344.61538461538464$$

$$a_4=a_1·r^{n-1}=910\cdot {0.6153846153846154}^{4-1}=910\cdot {0.6153846153846154}^3=910\cdot 0.23304506144742834=212.0710059171598$$

$$a_5=a_1·r^{n-1}=910\cdot {0.6153846153846154}^{5-1}=910\cdot {0.6153846153846154}^4=910\cdot 0.14341234550610976=130.5052344105599$$

$$a_6=a_1·r^{n-1}=910\cdot {0.6153846153846154}^{6-1}=910\cdot {0.6153846153846154}^5=910\cdot 0.08825375108068294=80.31091348342147$$

$$a_7=a_1·r^{n-1}=910\cdot {0.6153846153846154}^{7-1}=910\cdot {0.6153846153846154}^6=910\cdot 0.054310000665035656=49.42210060518244$$

$$a_8=a_1·r^{n-1}=910\cdot {0.6153846153846154}^{8-1}=910\cdot {0.6153846153846154}^7=910\cdot 0.033421538870791176=30.41360037241997$$

$$a_9=a_1·r^{n-1}=910\cdot {0.6153846153846154}^{9-1}=910\cdot {0.6153846153846154}^8=910\cdot 0.0205671008435638=18.71606176764306$$

$$a_{10}=a_1·r^{n-1}=910\cdot {0.6153846153846154}^{10-1}=910\cdot {0.6153846153846154}^9=910\cdot 0.012656677442193108=11.517576472395728$$