সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=80$$ এবং সাধারণ অনুপাত: $$r=1.2$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=80$$

$$a_2=a_1·r^{n-1}=80\cdot {1.2}^{2-1}=80\cdot {1.2}^1=80\cdot 1.2=96$$

$$a_3=a_1·r^{n-1}=80\cdot {1.2}^{3-1}=80\cdot {1.2}^2=80\cdot 1.44=115.19999999999999$$

$$a_4=a_1·r^{n-1}=80\cdot {1.2}^{4-1}=80\cdot {1.2}^3=80\cdot 1.7279999999999998=138.23999999999998$$

$$a_5=a_1·r^{n-1}=80\cdot {1.2}^{5-1}=80\cdot {1.2}^4=80\cdot 2.0736=165.88799999999998$$

$$a_6=a_1·r^{n-1}=80\cdot {1.2}^{6-1}=80\cdot {1.2}^5=80\cdot 2.4883199999999994=199.06559999999996$$

$$a_7=a_1·r^{n-1}=80\cdot {1.2}^{7-1}=80\cdot {1.2}^6=80\cdot 2.9859839999999993=238.87871999999993$$

$$a_8=a_1·r^{n-1}=80\cdot {1.2}^{8-1}=80\cdot {1.2}^7=80\cdot 3.583180799999999=286.65446399999996$$

$$a_9=a_1·r^{n-1}=80\cdot {1.2}^{9-1}=80\cdot {1.2}^8=80\cdot 4.2998169599999985=343.98535679999986$$

$$a_{10}=a_1·r^{n-1}=80\cdot {1.2}^{10-1}=80\cdot {1.2}^9=80\cdot 5.1597803519999985=412.7824281599999$$