সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=160$$ এবং সাধারণ অনুপাত: $$r=4.2$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=160$$

$$a_2=a_1·r^{n-1}=160\cdot {4.2}^{2-1}=160\cdot {4.2}^1=160\cdot 4.2=672$$

$$a_3=a_1·r^{n-1}=160\cdot {4.2}^{3-1}=160\cdot {4.2}^2=160\cdot 17.64=2822.4$$

$$a_4=a_1·r^{n-1}=160\cdot {4.2}^{4-1}=160\cdot {4.2}^3=160\cdot 74.08800000000001=11854.080000000002$$

$$a_5=a_1·r^{n-1}=160\cdot {4.2}^{5-1}=160\cdot {4.2}^4=160\cdot 311.16960000000006=49787.13600000001$$

$$a_6=a_1·r^{n-1}=160\cdot {4.2}^{6-1}=160\cdot {4.2}^5=160\cdot 1306.9123200000004=209105.97120000006$$

$$a_7=a_1·r^{n-1}=160\cdot {4.2}^{7-1}=160\cdot {4.2}^6=160\cdot 5489.031744000002=878245.0790400002$$

$$a_8=a_1·r^{n-1}=160\cdot {4.2}^{8-1}=160\cdot {4.2}^7=160\cdot 23053.933324800008=3688629.331968001$$

$$a_9=a_1·r^{n-1}=160\cdot {4.2}^{9-1}=160\cdot {4.2}^8=160\cdot 96826.51996416003=15492243.194265606$$

$$a_{10}=a_1·r^{n-1}=160\cdot {4.2}^{10-1}=160\cdot {4.2}^9=160\cdot 406671.38384947216=65067421.41591555$$