সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=100$$ এবং সাধারণ অনুপাত: $$r=0.02$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=100$$

$$a_2=a_1·r^{n-1}=100\cdot {0.02}^{2-1}=100\cdot {0.02}^1=100\cdot 0.02=2$$

$$a_3=a_1·r^{n-1}=100\cdot {0.02}^{3-1}=100\cdot {0.02}^2=100\cdot 0.0004=0.04$$

$$a_4=a_1·r^{n-1}=100\cdot {0.02}^{4-1}=100\cdot {0.02}^3=100\cdot 8.000000000000001E-06=0.0008000000000000001$$

$$a_5=a_1·r^{n-1}=100\cdot {0.02}^{5-1}=100\cdot {0.02}^4=100\cdot 1.6E-07=1.6E-05$$

$$a_6=a_1·r^{n-1}=100\cdot {0.02}^{6-1}=100\cdot {0.02}^5=100\cdot 3.2000000000000005E-09=3.2000000000000006E-07$$

$$a_7=a_1·r^{n-1}=100\cdot {0.02}^{7-1}=100\cdot {0.02}^6=100\cdot 6.400000000000001E-11=6.400000000000001E-09$$

$$a_8=a_1·r^{n-1}=100\cdot {0.02}^{8-1}=100\cdot {0.02}^7=100\cdot 1.2800000000000002E-12=1.2800000000000002E-10$$

$$a_9=a_1·r^{n-1}=100\cdot {0.02}^{9-1}=100\cdot {0.02}^8=100\cdot 2.5600000000000006E-14=2.5600000000000004E-12$$

$$a_{10}=a_1·r^{n-1}=100\cdot {0.02}^{10-1}=100\cdot {0.02}^9=100\cdot 5.120000000000001E-16=5.120000000000001E-14$$