সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=145$$ এবং সাধারণ অনুপাত: $$r=1.2$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=145$$

$$a_2=a_1·r^{n-1}=145\cdot {1.2}^{2-1}=145\cdot {1.2}^1=145\cdot 1.2=174$$

$$a_3=a_1·r^{n-1}=145\cdot {1.2}^{3-1}=145\cdot {1.2}^2=145\cdot 1.44=208.79999999999998$$

$$a_4=a_1·r^{n-1}=145\cdot {1.2}^{4-1}=145\cdot {1.2}^3=145\cdot 1.7279999999999998=250.55999999999997$$

$$a_5=a_1·r^{n-1}=145\cdot {1.2}^{5-1}=145\cdot {1.2}^4=145\cdot 2.0736=300.67199999999997$$

$$a_6=a_1·r^{n-1}=145\cdot {1.2}^{6-1}=145\cdot {1.2}^5=145\cdot 2.4883199999999994=360.80639999999994$$

$$a_7=a_1·r^{n-1}=145\cdot {1.2}^{7-1}=145\cdot {1.2}^6=145\cdot 2.9859839999999993=432.9676799999999$$

$$a_8=a_1·r^{n-1}=145\cdot {1.2}^{8-1}=145\cdot {1.2}^7=145\cdot 3.583180799999999=519.5612159999998$$

$$a_9=a_1·r^{n-1}=145\cdot {1.2}^{9-1}=145\cdot {1.2}^8=145\cdot 4.2998169599999985=623.4734591999998$$

$$a_{10}=a_1·r^{n-1}=145\cdot {1.2}^{10-1}=145\cdot {1.2}^9=145\cdot 5.1597803519999985=748.1681510399998$$