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Home » Find the sum of the following geometric series: (i) to 8 terms; (ii) to 8 terms;
Find the sum of the following geometric series:
(i) 0.15+0.015+0.0015+\ldots to 8 terms;
(ii) \sqrt{2}+1 / \sqrt{2}+1 / 2 \sqrt{2}+\ldots to 8 terms;
CBSE | Class 11 | Exercise 20.3 | Geometric Progressions | Maths | RD Sharma
Find the sum of the following geometric series:
(i) 0.15+0.015+0.0015+\ldots to 8 terms;
(ii) \sqrt{2}+1 / \sqrt{2}+1 / 2 \sqrt{2}+\ldots to 8 terms;

Solution:

(i) 0.15+0.015+0.0015+\ldots to 8 terms
Given that
\begin{array}{l} a=0.15 \\ r=t_{2} / t_{1}=0.015 / 0.15=0.1=1 / 10 \\ n=8 \end{array}
Using the formula,
The sum of GP for n terms =a\left(1-r^{n}\right) /(1-r)
\begin{array}{l} \mathrm{a}\left(1-\mathrm{r}^{\mathrm{n}}\right) /(1-\mathrm{r})=0.15\left(1-(1 / 10)^{8}\right) /(1-(1 / 10)) \\ =0.15\left(1-1 / 10^{8}\right) /(1 / 10) \\ =1 / 6\left(1-1 / 10^{8}\right) \end{array}

(ii) \sqrt{2}+1 / \sqrt{2}+1 / 2 \sqrt{2}+\ldots to 8 terms;
Given that
\begin{array}{l} a=\sqrt{2} \\ r=t_{2} / t_{1}=(1 / \sqrt{2}) / \sqrt{2}=1 / 2 \\ n=8 \end{array}
Using the formula,
The sum of GP for n terms =a\left(1-r^{n}\right) /(1-r)
\begin{array}{l} \mathrm{a}\left(1-\mathrm{r}^{\mathrm{n}}\right) /(1-\mathrm{r})=\sqrt{2}\left(1-(1 / 2)^{8}\right) /(1-(1 / 2)) \\ =\sqrt{2}(1-1 / 256) /(1 / 2) \\ =\sqrt{2}((256-1) / 256) \times 2 \\ =\sqrt{2}(255 \times 2) / 256 \\ =(255 \sqrt{2}) / 128 \end{array}

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