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Question Number 58364 by Hassen_Timol last updated on 22/Apr/19

$$\mathrm{Let}{a}_n=10\times {\left(\frac{1}{\sqrt{2}}\right)}^n$$$$a_n\mathrm{is}\mathrm{a}\mathrm{geometrical}\mathrm{sequence}$$$$S_n=a_0+a_1+...+a_{n-1}$$$$S_n=10\times \frac{1-{\left(\frac{1}{\sqrt{2}}\right)}^n}{1-\left(\frac{1}{\sqrt{2}}\right)}$$$$\mathbf{Proove}\mathbf{that}:$$$$$$$$S_n=\frac{10\sqrt{2}}{\sqrt{2}-1}\times (1-{\left(\frac{1}{\sqrt{2}}\right)}^n)$$$$$$

Answered by Kunal12588 last updated on 22/Apr/19

$$S_n=10\times \frac{1-{\left(\frac{1}{\sqrt{2}}\right)}^n}{1-\frac{1}{\sqrt{2}}}\left[Given\right]$$$$=10\sqrt{2}\times \frac{1-{\left(\frac{1}{\sqrt{2}}\right)}^n}{\sqrt{2}-1}$$$$=\frac{10\sqrt{2}}{\sqrt{2}-1}(1-{\left(\frac{1}{\sqrt{2}}\right)}^n)$$$$wellWhatwasthequestion?$$

Commented by Hassen_Timol last updated on 22/Apr/19

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}$$

Answered by tanmay last updated on 22/Apr/19

$$a_n=10\times {\left(\frac{1}{\sqrt{2}}\right)}^n\to a_1=10{\left(\frac{1}{\sqrt{2}}\right)}^1$$$$a_2=10\times {\left(\frac{1}{\sqrt{2}}\right)}^2$$$$formula{S}_n=\frac{a(1-r^n)}{1-r}$$$$r=\frac{a_2}{a_1}=\frac{10\times {\left(\frac{1}{\sqrt{2}}\right)}^2}{10\times {\left(\frac{1}{\sqrt{2}}\right)}^1}=\frac{1}{\sqrt{2}}$$$$S_n=\frac{a(1-r^n)}{1-r}$$$$=\frac{10\times \frac{1}{\sqrt{2}}\{(1-{\left(\frac{1}{\sqrt{2}}\right)}^n\}}{1-\frac{1}{\sqrt{2}}}$$$$=\frac{10\times \{(1-{\left(\frac{1}{\sqrt{2}}\right)}^n\}}{\sqrt{2}-1}$$

Commented by Hassen_Timol last updated on 22/Apr/19

$$\mathrm{Thank}\mathrm{you}\mathrm{Sir}$$

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