সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=1,950$$ এবং সাধারণ অনুপাত: $$r=0.7$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=1950$$

$$a_2=a_1·r^{n-1}=1950\cdot {0.7}^{2-1}=1950\cdot {0.7}^1=1950\cdot 0.7=1365$$

$$a_3=a_1·r^{n-1}=1950\cdot {0.7}^{3-1}=1950\cdot {0.7}^2=1950\cdot 0.48999999999999994=955.4999999999999$$

$$a_4=a_1·r^{n-1}=1950\cdot {0.7}^{4-1}=1950\cdot {0.7}^3=1950\cdot 0.3429999999999999=668.8499999999998$$

$$a_5=a_1·r^{n-1}=1950\cdot {0.7}^{5-1}=1950\cdot {0.7}^4=1950\cdot 0.24009999999999995=468.1949999999999$$

$$a_6=a_1·r^{n-1}=1950\cdot {0.7}^{6-1}=1950\cdot {0.7}^5=1950\cdot 0.16806999999999994=327.73649999999986$$

$$a_7=a_1·r^{n-1}=1950\cdot {0.7}^{7-1}=1950\cdot {0.7}^6=1950\cdot 0.11764899999999996=229.41554999999994$$

$$a_8=a_1·r^{n-1}=1950\cdot {0.7}^{8-1}=1950\cdot {0.7}^7=1950\cdot 0.08235429999999996=160.59088499999993$$

$$a_9=a_1·r^{n-1}=1950\cdot {0.7}^{9-1}=1950\cdot {0.7}^8=1950\cdot 0.05764800999999997=112.41361949999994$$

$$a_{10}=a_1·r^{n-1}=1950\cdot {0.7}^{10-1}=1950\cdot {0.7}^9=1950\cdot 0.04035360699999998=78.68953364999996$$