解答 - 几何数列

要找到系列的通用形式，将第一项：$$a=91$$ 和公比：$$r=1.2307692307692308$$ 插入几何级数的公式：

$$a_1=91$$

$$a_2=a_1·r^{n-1}=91\cdot {1.2307692307692308}^{2-1}=91\cdot {1.2307692307692308}^1=91\cdot 1.2307692307692308=112$$

$$a_3=a_1·r^{n-1}=91\cdot {1.2307692307692308}^{3-1}=91\cdot {1.2307692307692308}^2=91\cdot 1.5147928994082842=137.84615384615387$$

$$a_4=a_1·r^{n-1}=91\cdot {1.2307692307692308}^{4-1}=91\cdot {1.2307692307692308}^3=91\cdot 1.8643604915794267=169.65680473372782$$

$$a_5=a_1·r^{n-1}=91\cdot {1.2307692307692308}^{5-1}=91\cdot {1.2307692307692308}^4=91\cdot 2.294597528097756=208.8083750568958$$

$$a_6=a_1·r^{n-1}=91\cdot {1.2307692307692308}^{6-1}=91\cdot {1.2307692307692308}^5=91\cdot 2.824120034581854=256.99492314694874$$

$$a_7=a_1·r^{n-1}=91\cdot {1.2307692307692308}^{7-1}=91\cdot {1.2307692307692308}^6=91\cdot 3.475840042562282=316.30144387316767$$

$$a_8=a_1·r^{n-1}=91\cdot {1.2307692307692308}^{8-1}=91\cdot {1.2307692307692308}^7=91\cdot 4.2779569754612705=389.29408476697563$$

$$a_9=a_1·r^{n-1}=91\cdot {1.2307692307692308}^{9-1}=91\cdot {1.2307692307692308}^8=91\cdot 5.265177815952333=479.13118125166227$$

$$a_{10}=a_1·r^{n-1}=91\cdot {1.2307692307692308}^{10-1}=91\cdot {1.2307692307692308}^9=91\cdot 6.4802188504028715=589.6999153866614$$