সমাধান - জ্যামিতিক ধারা

সিরিজের সাধারণ রূপ খুঁজে পেতে, প্রথম পদ: $$a=95$$ এবং সাধারণ অনুপাত: $$r=1.1578947368421053$$ জ্যামিতিক সিরিজের সূত্রে প্লাগ করুন:

$$a_1=95$$

$$a_2=a_1·r^{n-1}=95\cdot {1.1578947368421053}^{2-1}=95\cdot {1.1578947368421053}^1=95\cdot 1.1578947368421053=110$$

$$a_3=a_1·r^{n-1}=95\cdot {1.1578947368421053}^{3-1}=95\cdot {1.1578947368421053}^2=95\cdot 1.3407202216066483=127.36842105263159$$

$$a_4=a_1·r^{n-1}=95\cdot {1.1578947368421053}^{4-1}=95\cdot {1.1578947368421053}^3=95\cdot 1.5524128881761192=147.47922437673134$$

$$a_5=a_1·r^{n-1}=95\cdot {1.1578947368421053}^{5-1}=95\cdot {1.1578947368421053}^4=95\cdot 1.7975307126249802=170.76541769937313$$

$$a_6=a_1·r^{n-1}=95\cdot {1.1578947368421053}^{6-1}=95\cdot {1.1578947368421053}^5=95\cdot 2.0813513514605035=197.72837838874784$$

$$a_7=a_1·r^{n-1}=95\cdot {1.1578947368421053}^{7-1}=95\cdot {1.1578947368421053}^6=95\cdot 2.4099857753753198=228.94864866065538$$

$$a_8=a_1·r^{n-1}=95\cdot {1.1578947368421053}^{8-1}=95\cdot {1.1578947368421053}^7=95\cdot 2.790509845171423=265.0984352912852$$

$$a_9=a_1·r^{n-1}=95\cdot {1.1578947368421053}^{9-1}=95\cdot {1.1578947368421053}^8=95\cdot 3.231116662830069=306.95608296885655$$

$$a_{10}=a_1·r^{n-1}=95\cdot {1.1578947368421053}^{10-1}=95\cdot {1.1578947368421053}^9=95\cdot 3.7412929780137643=355.4228329113076$$