সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=21,25,398$$ এবং সাধারণ অনুপাত: $$r=0.002409901580786281$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=2125398$$

$$a_2=a_1·r^{n-1}=2125398\cdot {0.002409901580786281}^{2-1}=2125398\cdot {0.002409901580786281}^1=2125398\cdot 0.002409901580786281=5122$$

$$a_3=a_1·r^{n-1}=2125398\cdot {0.002409901580786281}^{3-1}=2125398\cdot {0.002409901580786281}^2=2125398\cdot 5.807625629076216E-06=12.34351589678733$$

$$a_4=a_1·r^{n-1}=2125398\cdot {0.002409901580786281}^{4-1}=2125398\cdot {0.002409901580786281}^3=2125398\cdot 1.3995806184125692E-08=0.029746658472128377$$

$$a_5=a_1·r^{n-1}=2125398\cdot {0.002409901580786281}^{5-1}=2125398\cdot {0.002409901580786281}^4=2125398\cdot 3.3728515447502913E-11=7.16865192750918E-05$$

$$a_6=a_1·r^{n-1}=2125398\cdot {0.002409901580786281}^{6-1}=2125398\cdot {0.002409901580786281}^5=2125398\cdot 8.128240269451176E-14=1.727574561221099E-07$$

$$a_7=a_1·r^{n-1}=2125398\cdot {0.002409901580786281}^{7-1}=2125398\cdot {0.002409901580786281}^6=2125398\cdot 1.9588259074361095E-16=4.163284666012892E-10$$

$$a_8=a_1·r^{n-1}=2125398\cdot {0.002409901580786281}^{8-1}=2125398\cdot {0.002409901580786281}^7=2125398\cdot 4.720577650815401E-19=1.0033106297887752E-12$$

$$a_9=a_1·r^{n-1}=2125398\cdot {0.002409901580786281}^{9-1}=2125398\cdot {0.002409901580786281}^8=2125398\cdot 1.1376127542924425E-21=2.4178798727476485E-15$$

$$a_{10}=a_1·r^{n-1}=2125398\cdot {0.002409901580786281}^{10-1}=2125398\cdot {0.002409901580786281}^9=2125398\cdot 2.741534774891992E-24=5.82685252748589E-18$$