সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=1,024$$ এবং সাধারণ অনুপাত: $$r=0.25$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=1024$$

$$a_2=a_1·r^{n-1}=1024\cdot {0.25}^{2-1}=1024\cdot {0.25}^1=1024\cdot 0.25=256$$

$$a_3=a_1·r^{n-1}=1024\cdot {0.25}^{3-1}=1024\cdot {0.25}^2=1024\cdot 0.0625=64$$

$$a_4=a_1·r^{n-1}=1024\cdot {0.25}^{4-1}=1024\cdot {0.25}^3=1024\cdot 0.015625=16$$

$$a_5=a_1·r^{n-1}=1024\cdot {0.25}^{5-1}=1024\cdot {0.25}^4=1024\cdot 0.00390625=4$$

$$a_6=a_1·r^{n-1}=1024\cdot {0.25}^{6-1}=1024\cdot {0.25}^5=1024\cdot 0.0009765625=1$$

$$a_7=a_1·r^{n-1}=1024\cdot {0.25}^{7-1}=1024\cdot {0.25}^6=1024\cdot 0.000244140625=0.25$$

$$a_8=a_1·r^{n-1}=1024\cdot {0.25}^{8-1}=1024\cdot {0.25}^7=1024\cdot 6.103515625E-05=0.0625$$

$$a_9=a_1·r^{n-1}=1024\cdot {0.25}^{9-1}=1024\cdot {0.25}^8=1024\cdot 1.52587890625E-05=0.015625$$

$$a_{10}=a_1·r^{n-1}=1024\cdot {0.25}^{10-1}=1024\cdot {0.25}^9=1024\cdot 3.814697265625E-06=0.00390625$$