解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=42$$ 和公比：$$r=0.8095238095238095$$ 插入几何级数的公式：

$$a_1=42$$

$$a_2=a_1·r^{n-1}=42\cdot {0.8095238095238095}^{2-1}=42\cdot {0.8095238095238095}^1=42\cdot 0.8095238095238095=34$$

$$a_3=a_1·r^{n-1}=42\cdot {0.8095238095238095}^{3-1}=42\cdot {0.8095238095238095}^2=42\cdot 0.655328798185941=27.523809523809522$$

$$a_4=a_1·r^{n-1}=42\cdot {0.8095238095238095}^{4-1}=42\cdot {0.8095238095238095}^3=42\cdot 0.5305042651981428=22.281179138321995$$

$$a_5=a_1·r^{n-1}=42\cdot {0.8095238095238095}^{5-1}=42\cdot {0.8095238095238095}^4=42\cdot 0.4294558337318299=18.037145016736854$$

$$a_6=a_1·r^{n-1}=42\cdot {0.8095238095238095}^{6-1}=42\cdot {0.8095238095238095}^5=42\cdot 0.34765472254481466=14.601498346882217$$

$$a_7=a_1·r^{n-1}=42\cdot {0.8095238095238095}^{7-1}=42\cdot {0.8095238095238095}^6=42\cdot 0.2814347753934214=11.820260566523698$$

$$a_8=a_1·r^{n-1}=42\cdot {0.8095238095238095}^{8-1}=42\cdot {0.8095238095238095}^7=42\cdot 0.22782815150896019=9.568782363376329$$

$$a_9=a_1·r^{n-1}=42\cdot {0.8095238095238095}^{9-1}=42\cdot {0.8095238095238095}^8=42\cdot 0.1844323131263011=7.746157151304646$$

$$a_{10}=a_1·r^{n-1}=42\cdot {0.8095238095238095}^{10-1}=42\cdot {0.8095238095238095}^9=42\cdot 0.14930234872129136=6.270698646294237$$