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Geometric Progressions — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSGeometric Progressions3 Marks
Question
If S denotes the sum of an infinite G.P. and $S_1$ denotes the sum of the squares of its terms, then prove that the first terms and common ratio are respectively $\frac{2\text{SS}_1}{\text{S}^2+\text{S}_1}\text{ and }\frac{\text{S}^2-\text{S}_1}{\text{S}^2+\text{S}_1}.$
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Answer

$\text{S}=\text{a}+\text{ar}+\text{ar}^2+\text{ar}^3+\ \dots$ $\text{S}=\frac{\text{a}}{1-\text{r}}\ \cdots(1)$ $\text{S}_1=\text{a}^2+\text{a}^2\text{r}^2+\text{a}^2\text{r}^4+\text{a}^2\text{r}^6\ \dots$
$\text{S}_1=\frac{\text{a}^2}{1-\text{r}^2}\ \dots(2)$ $\text{S}^2=\frac{\text{a}^2}{(1-\text{r})^2}$
$\text{S}^2=\frac{\text{S}_1\big(1-\text{r}^2\big)}{\big(1-\text{r}^2\big)}$
$\big(1-\text{r}\big)\text{S}^2=\text{S}_1(1+\text{r})$
$\text{S}^2-\text{S}^2\text{r}=\text{S}_1-\text{S}_1\text{r}$
$\text{S}_1\text{r}+\text{S}^2\text{r}=\text{S}^2-\text{S}_1$
$\text{r}=\frac{\text{S}^2-\text{S}_1}{\text{S}_1+\text{S}^2}$ Put r in equition (1) $\text{S}(1-\text{r})=\text{a}$
$\text{a}=\text{S}\bigg[1-\frac{\text{S}^2-\text{S}_1}{\text{S}^2+\text{S}_1}\bigg]$
$\text{a}=\text{S}\bigg[\frac{\text{S}^2+\text{S}_1-\text{S}^2+\text{S}_1}{\text{S}^2+\text{S}_1}\bigg]$ $\text{a}=\frac{2\text{S}\text{S}_1}{\text{S}^2+\text{S}_1}$

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