সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=90$$ এবং সাধারণ অনুপাত: $$r=1.2$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=90$$

$$a_2=a_1·r^{n-1}=90\cdot {1.2}^{2-1}=90\cdot {1.2}^1=90\cdot 1.2=108$$

$$a_3=a_1·r^{n-1}=90\cdot {1.2}^{3-1}=90\cdot {1.2}^2=90\cdot 1.44=129.6$$

$$a_4=a_1·r^{n-1}=90\cdot {1.2}^{4-1}=90\cdot {1.2}^3=90\cdot 1.7279999999999998=155.51999999999998$$

$$a_5=a_1·r^{n-1}=90\cdot {1.2}^{5-1}=90\cdot {1.2}^4=90\cdot 2.0736=186.624$$

$$a_6=a_1·r^{n-1}=90\cdot {1.2}^{6-1}=90\cdot {1.2}^5=90\cdot 2.4883199999999994=223.94879999999995$$

$$a_7=a_1·r^{n-1}=90\cdot {1.2}^{7-1}=90\cdot {1.2}^6=90\cdot 2.9859839999999993=268.73855999999995$$

$$a_8=a_1·r^{n-1}=90\cdot {1.2}^{8-1}=90\cdot {1.2}^7=90\cdot 3.583180799999999=322.48627199999993$$

$$a_9=a_1·r^{n-1}=90\cdot {1.2}^{9-1}=90\cdot {1.2}^8=90\cdot 4.2998169599999985=386.98352639999985$$

$$a_{10}=a_1·r^{n-1}=90\cdot {1.2}^{10-1}=90\cdot {1.2}^9=90\cdot 5.1597803519999985=464.38023167999984$$