সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=65$$ এবং সাধারণ অনুপাত: $$r=1.2$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=65$$

$$a_2=a_1·r^{n-1}=65\cdot {1.2}^{2-1}=65\cdot {1.2}^1=65\cdot 1.2=78$$

$$a_3=a_1·r^{n-1}=65\cdot {1.2}^{3-1}=65\cdot {1.2}^2=65\cdot 1.44=93.6$$

$$a_4=a_1·r^{n-1}=65\cdot {1.2}^{4-1}=65\cdot {1.2}^3=65\cdot 1.7279999999999998=112.31999999999998$$

$$a_5=a_1·r^{n-1}=65\cdot {1.2}^{5-1}=65\cdot {1.2}^4=65\cdot 2.0736=134.784$$

$$a_6=a_1·r^{n-1}=65\cdot {1.2}^{6-1}=65\cdot {1.2}^5=65\cdot 2.4883199999999994=161.74079999999995$$

$$a_7=a_1·r^{n-1}=65\cdot {1.2}^{7-1}=65\cdot {1.2}^6=65\cdot 2.9859839999999993=194.08895999999996$$

$$a_8=a_1·r^{n-1}=65\cdot {1.2}^{8-1}=65\cdot {1.2}^7=65\cdot 3.583180799999999=232.90675199999995$$

$$a_9=a_1·r^{n-1}=65\cdot {1.2}^{9-1}=65\cdot {1.2}^8=65\cdot 4.2998169599999985=279.4881023999999$$

$$a_{10}=a_1·r^{n-1}=65\cdot {1.2}^{10-1}=65\cdot {1.2}^9=65\cdot 5.1597803519999985=335.3857228799999$$