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Geometric Progressions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsGeometric Progressions3 Marks
Question
Find the sum of the following geometric progrssions:$(\text{a}^2-\text{b}^2),(\text{a}-\text{b}),\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\Big),\ ...\text{to n terms}$
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Answer

$(\text{a}^2-\text{b}^2),(\text{a}-\text{b}),\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\Big), ...\text{n terms}$$\text{a}=\text{a}^2-\text{b}^2,\text{r}=\frac{\text{a}-\text{b}}{\text{a}^2-\text{b}^2}=\frac{1}{\text{a}+\text{b}},\text{n}=\text{n}$
$\text{S}_\text{n}=\text{a}\frac{(\text{r}^{1-\text{n}})}{\text{r}-1}$ $[\because\text{r}<1]$
$\text{S}_\text{n}=(\text{a}^2-\text{b}^2)\frac{\big(1-\frac{1}{(\text{a}+\text{b})^\text{n}}\big)}{1-\frac{1}{\text{a}+\text{b}}}$
$=\frac{(\text{a}-\text{b})((\text{a}+\text{b})^\text{n}-1)}{\frac{(\text{a}+\text{b})^{-1}(\text{a}+\text{b})^\text{n}(\text{a}+\text{b})-1}{(\text{a}+\text{b})}}$
$=\frac{\text{a}-\text{b}}{{(\text{a}+\text{b})^{\text{n}-2}}}\Big\{\frac{(\text{a}+\text{b})^\text{n}-1}{(\text{a}+\text{b})-1}\Big\}$

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