সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=120$$ এবং সাধারণ অনুপাত: $$r=0.8$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=120$$

$$a_2=a_1·r^{n-1}=120\cdot {0.8}^{2-1}=120\cdot {0.8}^1=120\cdot 0.8=96$$

$$a_3=a_1·r^{n-1}=120\cdot {0.8}^{3-1}=120\cdot {0.8}^2=120\cdot 0.6400000000000001=76.80000000000001$$

$$a_4=a_1·r^{n-1}=120\cdot {0.8}^{4-1}=120\cdot {0.8}^3=120\cdot 0.5120000000000001=61.44000000000001$$

$$a_5=a_1·r^{n-1}=120\cdot {0.8}^{5-1}=120\cdot {0.8}^4=120\cdot 0.4096000000000001=49.15200000000001$$

$$a_6=a_1·r^{n-1}=120\cdot {0.8}^{6-1}=120\cdot {0.8}^5=120\cdot 0.3276800000000001=39.32160000000001$$

$$a_7=a_1·r^{n-1}=120\cdot {0.8}^{7-1}=120\cdot {0.8}^6=120\cdot 0.2621440000000001=31.45728000000001$$

$$a_8=a_1·r^{n-1}=120\cdot {0.8}^{8-1}=120\cdot {0.8}^7=120\cdot 0.20971520000000007=25.165824000000008$$

$$a_9=a_1·r^{n-1}=120\cdot {0.8}^{9-1}=120\cdot {0.8}^8=120\cdot 0.1677721600000001=20.13265920000001$$

$$a_{10}=a_1·r^{n-1}=120\cdot {0.8}^{10-1}=120\cdot {0.8}^9=120\cdot 0.13421772800000006=16.10612736000001$$