সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=120$$ এবং সাধারণ অনুপাত: $$r=1.6$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=120$$

$$a_2=a_1·r^{n-1}=120\cdot {1.6}^{2-1}=120\cdot {1.6}^1=120\cdot 1.6=192$$

$$a_3=a_1·r^{n-1}=120\cdot {1.6}^{3-1}=120\cdot {1.6}^2=120\cdot 2.5600000000000005=307.20000000000005$$

$$a_4=a_1·r^{n-1}=120\cdot {1.6}^{4-1}=120\cdot {1.6}^3=120\cdot 4.096000000000001=491.5200000000001$$

$$a_5=a_1·r^{n-1}=120\cdot {1.6}^{5-1}=120\cdot {1.6}^4=120\cdot 6.553600000000001=786.4320000000001$$

$$a_6=a_1·r^{n-1}=120\cdot {1.6}^{6-1}=120\cdot {1.6}^5=120\cdot 10.485760000000003=1258.2912000000003$$

$$a_7=a_1·r^{n-1}=120\cdot {1.6}^{7-1}=120\cdot {1.6}^6=120\cdot 16.777216000000006=2013.2659200000007$$

$$a_8=a_1·r^{n-1}=120\cdot {1.6}^{8-1}=120\cdot {1.6}^7=120\cdot 26.84354560000001=3221.225472000001$$

$$a_9=a_1·r^{n-1}=120\cdot {1.6}^{9-1}=120\cdot {1.6}^8=120\cdot 42.94967296000002=5153.9607552000025$$

$$a_{10}=a_1·r^{n-1}=120\cdot {1.6}^{10-1}=120\cdot {1.6}^9=120\cdot 68.71947673600003=8246.337208320005$$