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Question Number 56411 by gunawan last updated on 16/Mar/19

$$\mathrm{For}\mathrm{a}\mathrm{sequence}<a_n>;a_1=2,\frac{a_{n+1}}{a_n}=\frac{1}{3},$$ $$\mathrm{then}\underset{r=1}{\sum ^{20}}{a}_r\mathrm{is}\mathrm{equal}\mathrm{to}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Mar/19

$$\frac{a_2}{a_1}=\frac{1}{3}=\frac{a_2}{2}\to a_2=\frac{2}{3}$$ $$\frac{a_3}{a_2}=\frac{1}{3}=\frac{a_3}{\frac{2}{3}}\to a_3=\frac{2}{3^2}$$ $$\frac{a_4}{a_3}=\frac{1}{3}=\frac{a_4}{\frac{2}{3^2}}\to a_4=\frac{2}{3^3}$$ $$S_n=a_1+a_2+a_3+...+a_n$$ $$=\frac{2}{1}+\frac{2}{3}+\frac{2}{3^2}+...+\frac{2}{3^{n-1}}$$ $$A=2r=\frac{1}{3}$$ $$S_n=\frac{2(1-\frac{1}{3^n})}{1-\frac{1}{3}}\to \frac{2(1-\frac{1}{3^n})}{\frac{2}{3}}\to 3(1-\frac{1}{3^n})$$ $$so{s}_{20}=3(1-\frac{1}{3^{20}})$$ $$$$ $$$$ $$$$ $$$$

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