Geometric Progressions — MATHS STD 11 Science — Question

$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$

$\text{a}=\sqrt{2},\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=\frac{\text{t}_2}{\text{t}_1}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac12$

$\text{t}_\text{n}=\frac{1}{512\sqrt{2}},\text{n}=?$

$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$

$\frac{1}{5125\sqrt{2}}=\big(\sqrt{2}\big)\Big(\frac12\Big)^{\text{n}-1}$

$\frac{1}{512\times\sqrt{2}\times\sqrt{2}}=\Big(\frac{1}{2}\Big)^{\text{n}-1}$

$\frac{1}{1024}=\Big(\frac12\Big)^{\text{n}-1}$

$\Big(\frac12\Big)^{10}=\Big(\frac12\Big)^{{\text{n}-1}}$

$10=(\text{n}-1)$

$\text{n}=11$

$\therefore\text{term is 11}^\text{th}.$

$\sqrt{3},3,3\sqrt{3},\ \dots\text{is}729?$

$\text{a}=\sqrt{3},\text{r}=\frac{\text{t}_\text{n}}{\text{t}_{\text{n}-1}}=,\text{n}=?,\text{t}_\text{n}=729$

Now,

$\text{t}_\text{n}=\text{ar}^{\text{n}-1}$

$729=\big(\sqrt{3}\big)(\text{r})^{\text{n}-1}$

Now,

$\text{r}=\frac{\text{t}_2}{\text{t}_1}=\frac{3}{\sqrt{3}}=\sqrt{3}$

$729=\big(\sqrt{3}\big)\big(\sqrt{3}\big)^{\text{n}-1}$

$729=\Big(\sqrt{3}\Big)^\text{n}$

$(3)^6=\big(\sqrt{3}\big)^{\text{n}}$

$\big(\sqrt{3}\big)^{12}=\big(\sqrt{3}\big)^{\text{n}}$

$\Rightarrow{\text{n}=12}$

$\therefore12^{\text{th}}\text{ term is 729}.$