সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=1,216$$ এবং সাধারণ অনুপাত: $$r=0.02631578947368421$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=1216$$

$$a_2=a_1·r^{n-1}=1216\cdot {0.02631578947368421}^{2-1}=1216\cdot {0.02631578947368421}^1=1216\cdot 0.02631578947368421=32$$

$$a_3=a_1·r^{n-1}=1216\cdot {0.02631578947368421}^{3-1}=1216\cdot {0.02631578947368421}^2=1216\cdot 0.0006925207756232686=0.8421052631578946$$

$$a_4=a_1·r^{n-1}=1216\cdot {0.02631578947368421}^{4-1}=1216\cdot {0.02631578947368421}^3=1216\cdot 1.8224230937454436E-05=0.022160664819944595$$

$$a_5=a_1·r^{n-1}=1216\cdot {0.02631578947368421}^{5-1}=1216\cdot {0.02631578947368421}^4=1216\cdot 4.795850246698536E-07=0.0005831753899985419$$

$$a_6=a_1·r^{n-1}=1216\cdot {0.02631578947368421}^{6-1}=1216\cdot {0.02631578947368421}^5=1216\cdot 1.2620658543943515E-08=1.5346720789435315E-05$$

$$a_7=a_1·r^{n-1}=1216\cdot {0.02631578947368421}^{7-1}=1216\cdot {0.02631578947368421}^6=1216\cdot 3.3212259326167143E-10=4.0386107340619247E-07$$

$$a_8=a_1·r^{n-1}=1216\cdot {0.02631578947368421}^{8-1}=1216\cdot {0.02631578947368421}^7=1216\cdot 8.740068243728195E-12=1.0627922984373484E-08$$

$$a_9=a_1·r^{n-1}=1216\cdot {0.02631578947368421}^{9-1}=1216\cdot {0.02631578947368421}^8=1216\cdot 2.3000179588758405E-13=2.796821837993022E-10$$

$$a_{10}=a_1·r^{n-1}=1216\cdot {0.02631578947368421}^{10-1}=1216\cdot {0.02631578947368421}^9=1216\cdot 6.0526788391469485E-15=7.36005746840269E-12$$