সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=60$$ এবং সাধারণ অনুপাত: $$r=0.05$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=60$$

$$a_2=a_1·r^{n-1}=60\cdot {0.05}^{2-1}=60\cdot {0.05}^1=60\cdot 0.05=3$$

$$a_3=a_1·r^{n-1}=60\cdot {0.05}^{3-1}=60\cdot {0.05}^2=60\cdot 0.0025000000000000005=0.15000000000000002$$

$$a_4=a_1·r^{n-1}=60\cdot {0.05}^{4-1}=60\cdot {0.05}^3=60\cdot 0.00012500000000000003=0.0075000000000000015$$

$$a_5=a_1·r^{n-1}=60\cdot {0.05}^{5-1}=60\cdot {0.05}^4=60\cdot 6.250000000000001E-06=0.00037500000000000006$$

$$a_6=a_1·r^{n-1}=60\cdot {0.05}^{6-1}=60\cdot {0.05}^5=60\cdot 3.125000000000001E-07=1.8750000000000005E-05$$

$$a_7=a_1·r^{n-1}=60\cdot {0.05}^{7-1}=60\cdot {0.05}^6=60\cdot 1.5625000000000006E-08=9.375000000000003E-07$$

$$a_8=a_1·r^{n-1}=60\cdot {0.05}^{8-1}=60\cdot {0.05}^7=60\cdot 7.812500000000003E-10=4.6875000000000014E-08$$

$$a_9=a_1·r^{n-1}=60\cdot {0.05}^{9-1}=60\cdot {0.05}^8=60\cdot 3.906250000000002E-11=2.343750000000001E-09$$

$$a_{10}=a_1·r^{n-1}=60\cdot {0.05}^{10-1}=60\cdot {0.05}^9=60\cdot 1.953125000000001E-12=1.1718750000000007E-10$$