সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=95$$ এবং সাধারণ অনুপাত: $$r=3.2$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=95$$

$$a_2=a_1·r^{n-1}=95\cdot {3.2}^{2-1}=95\cdot {3.2}^1=95\cdot 3.2=304$$

$$a_3=a_1·r^{n-1}=95\cdot {3.2}^{3-1}=95\cdot {3.2}^2=95\cdot 10.240000000000002=972.8000000000002$$

$$a_4=a_1·r^{n-1}=95\cdot {3.2}^{4-1}=95\cdot {3.2}^3=95\cdot 32.76800000000001=3112.960000000001$$

$$a_5=a_1·r^{n-1}=95\cdot {3.2}^{5-1}=95\cdot {3.2}^4=95\cdot 104.85760000000002=9961.472000000002$$

$$a_6=a_1·r^{n-1}=95\cdot {3.2}^{6-1}=95\cdot {3.2}^5=95\cdot 335.5443200000001=31876.710400000007$$

$$a_7=a_1·r^{n-1}=95\cdot {3.2}^{7-1}=95\cdot {3.2}^6=95\cdot 1073.7418240000004=102005.47328000003$$

$$a_8=a_1·r^{n-1}=95\cdot {3.2}^{8-1}=95\cdot {3.2}^7=95\cdot 3435.973836800001=326417.51449600013$$

$$a_9=a_1·r^{n-1}=95\cdot {3.2}^{9-1}=95\cdot {3.2}^8=95\cdot 10995.116277760006=1044536.0463872006$$

$$a_{10}=a_1·r^{n-1}=95\cdot {3.2}^{10-1}=95\cdot {3.2}^9=95\cdot 35184.37208883202=3342515.3484390415$$