সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=100$$ এবং সাধারণ অনুপাত: $$r=0.12$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=100$$

$$a_2=a_1·r^{n-1}=100\cdot {0.12}^{2-1}=100\cdot {0.12}^1=100\cdot 0.12=12$$

$$a_3=a_1·r^{n-1}=100\cdot {0.12}^{3-1}=100\cdot {0.12}^2=100\cdot 0.0144=1.44$$

$$a_4=a_1·r^{n-1}=100\cdot {0.12}^{4-1}=100\cdot {0.12}^3=100\cdot 0.0017279999999999997=0.17279999999999998$$

$$a_5=a_1·r^{n-1}=100\cdot {0.12}^{5-1}=100\cdot {0.12}^4=100\cdot 0.00020735999999999997=0.020735999999999997$$

$$a_6=a_1·r^{n-1}=100\cdot {0.12}^{6-1}=100\cdot {0.12}^5=100\cdot 2.4883199999999996E-05=0.0024883199999999996$$

$$a_7=a_1·r^{n-1}=100\cdot {0.12}^{7-1}=100\cdot {0.12}^6=100\cdot 2.9859839999999994E-06=0.00029859839999999994$$

$$a_8=a_1·r^{n-1}=100\cdot {0.12}^{8-1}=100\cdot {0.12}^7=100\cdot 3.583180799999999E-07=3.583180799999999E-05$$

$$a_9=a_1·r^{n-1}=100\cdot {0.12}^{9-1}=100\cdot {0.12}^8=100\cdot 4.299816959999999E-08=4.299816959999999E-06$$

$$a_{10}=a_1·r^{n-1}=100\cdot {0.12}^{10-1}=100\cdot {0.12}^9=100\cdot 5.1597803519999985E-09=5.159780351999999E-07$$