সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=240$$ এবং সাধারণ অনুপাত: $$r=4.3$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=240$$

$$a_2=a_1·r^{n-1}=240\cdot {4.3}^{2-1}=240\cdot {4.3}^1=240\cdot 4.3=1032$$

$$a_3=a_1·r^{n-1}=240\cdot {4.3}^{3-1}=240\cdot {4.3}^2=240\cdot 18.49=4437.599999999999$$

$$a_4=a_1·r^{n-1}=240\cdot {4.3}^{4-1}=240\cdot {4.3}^3=240\cdot 79.50699999999999=19081.679999999997$$

$$a_5=a_1·r^{n-1}=240\cdot {4.3}^{5-1}=240\cdot {4.3}^4=240\cdot 341.88009999999997=82051.22399999999$$

$$a_6=a_1·r^{n-1}=240\cdot {4.3}^{6-1}=240\cdot {4.3}^5=240\cdot 1470.0844299999997=352820.2631999999$$

$$a_7=a_1·r^{n-1}=240\cdot {4.3}^{7-1}=240\cdot {4.3}^6=240\cdot 6321.363048999999=1517127.1317599998$$

$$a_8=a_1·r^{n-1}=240\cdot {4.3}^{8-1}=240\cdot {4.3}^7=240\cdot 27181.861110699992=6523646.666567998$$

$$a_9=a_1·r^{n-1}=240\cdot {4.3}^{9-1}=240\cdot {4.3}^8=240\cdot 116882.00277600996=28051680.66624239$$

$$a_{10}=a_1·r^{n-1}=240\cdot {4.3}^{10-1}=240\cdot {4.3}^9=240\cdot 502592.6119368428=120622226.86484227$$