सोल्यूशन - भूमितीय अनुक्रम

मालिकेचा सामान्य रूप कसा असेल हे शोधण्यासाठी, पहिला मूळभूत: $$a=42$$ आणि सामान्य अनुपात: $$r=1.119047619047619$$ या भूतगणितीय मालिकेच्या सूत्रात ठेवा:

$$a_1=42$$

$$a_2=a_1·r^{n-1}=42\cdot {1.119047619047619}^{2-1}=42\cdot {1.119047619047619}^1=42\cdot 1.119047619047619=47$$

$$a_3=a_1·r^{n-1}=42\cdot {1.119047619047619}^{3-1}=42\cdot {1.119047619047619}^2=42\cdot 1.2522675736961453=52.5952380952381$$

$$a_4=a_1·r^{n-1}=42\cdot {1.119047619047619}^{4-1}=42\cdot {1.119047619047619}^3=42\cdot 1.40134704675521=58.856575963718825$$

$$a_5=a_1·r^{n-1}=42\cdot {1.119047619047619}^{5-1}=42\cdot {1.119047619047619}^4=42\cdot 1.5681740761308305=65.86331119749488$$

$$a_6=a_1·r^{n-1}=42\cdot {1.119047619047619}^{6-1}=42\cdot {1.119047619047619}^5=42\cdot 1.7548614661464055=73.70418157814903$$

$$a_7=a_1·r^{n-1}=42\cdot {1.119047619047619}^{7-1}=42\cdot {1.119047619047619}^6=42\cdot 1.963773545449549=82.47848890888106$$

$$a_8=a_1·r^{n-1}=42\cdot {1.119047619047619}^{8-1}=42\cdot {1.119047619047619}^7=42\cdot 2.1975561103840193=92.29735663612881$$

$$a_9=a_1·r^{n-1}=42\cdot {1.119047619047619}^{9-1}=42\cdot {1.119047619047619}^8=42\cdot 2.4591699330487833=103.28513718804889$$

$$a_{10}=a_1·r^{n-1}=42\cdot {1.119047619047619}^{10-1}=42\cdot {1.119047619047619}^9=42\cdot 2.7519282584117337=115.58098685329281$$