সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=1,001$$ এবং সাধারণ অনুপাত: $$r=0.01998001998001998$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=1001$$

$$a_2=a_1·r^{n-1}=1001\cdot {0.01998001998001998}^{2-1}=1001\cdot {0.01998001998001998}^1=1001\cdot 0.01998001998001998=20$$

$$a_3=a_1·r^{n-1}=1001\cdot {0.01998001998001998}^{3-1}=1001\cdot {0.01998001998001998}^2=1001\cdot 0.0003992011984019976=0.3996003996003996$$

$$a_4=a_1·r^{n-1}=1001\cdot {0.01998001998001998}^{4-1}=1001\cdot {0.01998001998001998}^3=1001\cdot 7.976047920119832E-06=0.007984023968039952$$

$$a_5=a_1·r^{n-1}=1001\cdot {0.01998001998001998}^{5-1}=1001\cdot {0.01998001998001998}^4=1001\cdot 1.5936159680559105E-07=0.00015952095840239663$$

$$a_6=a_1·r^{n-1}=1001\cdot {0.01998001998001998}^{6-1}=1001\cdot {0.01998001998001998}^5=1001\cdot 3.1840478882235973E-09=3.187231936111821E-06$$

$$a_7=a_1·r^{n-1}=1001\cdot {0.01998001998001998}^{7-1}=1001\cdot {0.01998001998001998}^6=1001\cdot 6.36173404240479E-11=6.368095776447195E-08$$

$$a_8=a_1·r^{n-1}=1001\cdot {0.01998001998001998}^{8-1}=1001\cdot {0.01998001998001998}^7=1001\cdot 1.2710757327482097E-12=1.272346808480958E-09$$

$$a_9=a_1·r^{n-1}=1001\cdot {0.01998001998001998}^{9-1}=1001\cdot {0.01998001998001998}^8=1001\cdot 2.539611853642777E-14=2.5421514654964197E-11$$

$$a_{10}=a_1·r^{n-1}=1001\cdot {0.01998001998001998}^{10-1}=1001\cdot {0.01998001998001998}^9=1001\cdot 5.074149557727826E-16=5.079223707285554E-13$$