সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=95$$ এবং সাধারণ অনুপাত: $$r=0.6$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=95$$

$$a_2=a_1·r^{n-1}=95\cdot {0.6}^{2-1}=95\cdot {0.6}^1=95\cdot 0.6=57$$

$$a_3=a_1·r^{n-1}=95\cdot {0.6}^{3-1}=95\cdot {0.6}^2=95\cdot 0.36=34.199999999999996$$

$$a_4=a_1·r^{n-1}=95\cdot {0.6}^{4-1}=95\cdot {0.6}^3=95\cdot 0.21599999999999997=20.519999999999996$$

$$a_5=a_1·r^{n-1}=95\cdot {0.6}^{5-1}=95\cdot {0.6}^4=95\cdot 0.1296=12.312$$

$$a_6=a_1·r^{n-1}=95\cdot {0.6}^{6-1}=95\cdot {0.6}^5=95\cdot 0.07775999999999998=7.387199999999998$$

$$a_7=a_1·r^{n-1}=95\cdot {0.6}^{7-1}=95\cdot {0.6}^6=95\cdot 0.04665599999999999=4.432319999999999$$

$$a_8=a_1·r^{n-1}=95\cdot {0.6}^{8-1}=95\cdot {0.6}^7=95\cdot 0.027993599999999993=2.6593919999999995$$

$$a_9=a_1·r^{n-1}=95\cdot {0.6}^{9-1}=95\cdot {0.6}^8=95\cdot 0.016796159999999994=1.5956351999999994$$

$$a_{10}=a_1·r^{n-1}=95\cdot {0.6}^{10-1}=95\cdot {0.6}^9=95\cdot 0.010077695999999997=0.9573811199999998$$