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Geometric Progressions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsGeometric Progressions3 Marks
Question
Which term of the G.P.:$\sqrt{3},3,3\sqrt{3},\ \dots\text{is}729?$
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Answer

$\sqrt{2},\frac{1}{\sqrt{2}},\frac{1}{2\sqrt{2}},\frac{1}{4\sqrt{2}},\ \dots\text{is}\frac{1}{512\sqrt{2}}?$$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\text{a}=\sqrt{2},\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=\frac{\text{t}_2}{\text{t}_1}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac12$
$\text{t}_\text{n}=\frac{1}{512\sqrt{2}},\text{n}=?$
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$\frac{1}{5125\sqrt{2}}=\big(\sqrt{2}\big)\Big(\frac12\Big)^{\text{n}-1}$
$\frac{1}{512\times\sqrt{2}\times\sqrt{2}}=\Big(\frac{1}{2}\Big)^{\text{n}-1}$
$\frac{1}{1024}=\Big(\frac12\Big)^{\text{n}-1}$
$\Big(\frac12\Big)^{10}=\Big(\frac12\Big)^{{\text{n}-1}}$
$10=(\text{n}-1)$
$\text{n}=11$
$\therefore\text{term is 11}^\text{th}.$
$\sqrt{3},3,3\sqrt{3},\ \dots\text{is}729?$
$\text{a}=\sqrt{3},\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=,\text{n}=?,\text{t}_\text{n}=729$
Now,
$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$
$729=\big(\sqrt{3}\big)(\text{r})^{\text{n}-1}$
Now,
$\text{r}=\frac{\text{t}_2}{\text{t}_1}=\frac{3}{\sqrt{3}}=\sqrt{3}$
$729=\big(\sqrt{3}\big)\big(\sqrt{3}\big)^{\text{n}-1}$
$729=\Big(\sqrt{3}\Big)^\text{n}$
$(3)^6=\big(\sqrt{3}\big)^{\text{n}}$
$\big(\sqrt{3}\big)^{12}=\big(\sqrt{3}\big)^{\text{n}}$
$\Rightarrow{\text{n}=12}$
$\therefore12^{\text{th}}\text{ term is 729}.$

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