Question

Which term of the G.P. :

(i) $\sqrt{2},\frac{1}{\sqrt{2}},\frac{1}{2\sqrt{2}},\frac{1}{4\sqrt{2}},....is\frac{1}{512\sqrt{2}}$ ?

(ii) 2, $2\sqrt{2},4,....is128$ ?

(iii) $\sqrt{3},3,3\sqrt{3},.....is729$ ?

(iv) $\frac{1}{3},\frac{1}{9},\frac{1}{27}...is\frac{1}{19683}$ ?

Answer 1

(i) $\sqrt{2},\frac{1}{\sqrt{2}},\frac{1}{2\sqrt{2}},\frac{1}{4\sqrt{2}},...$ is $\frac{1}{512\sqrt{2}}$

$t_n=a{r}^{n-1}$

$a=\sqrt{2},r=\frac{t_n}{t_{n-1}}=\frac{t_2}{t_1}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac{1}{2}$

$t_n=\frac{1}{512\sqrt{2}},n=?$

$t_n=a{r}^{n-1}$

$\Rightarrow \frac{1}{512\sqrt{2}}=\left(\sqrt{2}\right){\left(\frac{1}{2}\right)}^{n-1}$

$\Rightarrow \frac{1}{512\times \sqrt{2}\times \sqrt{2}}={\left(\frac{1}{2}\right)}^{n-1}$

$\Rightarrow \frac{1}{1024}={\left(\frac{1}{2}\right)}^{n-1}$

$\Rightarrow {\left(\frac{1}{2}\right)}^{10}={\left(\frac{1}{2}\right)}^{n-1}$

$\Rightarrow 10=(n-1)$

$\Rightarrow n=11$

Thus, the ${11}^{th}$ term of the given G.P. is $\frac{1}{512\sqrt{2}}$.

(ii) $2,\sqrt{2},4,....is128$

$a=2,r=\frac{t_n}{t_{n-1}}=\frac{2\sqrt{2}}{2}=\sqrt{2},n=?$

$t_n=128$

Also,

$\Rightarrow t_n=a{r}^{n-1}$

$\Rightarrow 128=\left(2\right)(\sqrt{2}{)}^{n-1}$

$\Rightarrow \frac{128}{2}=(\sqrt{2}{)}^{n-1}$

$\Rightarrow 64=(\sqrt{2}{)}^{n-1}$

$\Rightarrow (\sqrt{2}{)}^{12}=(\sqrt{2}{)}^{n-1}$

$\Rightarrow 12=n-1$

$\Rightarrow n=13$

Thus, the ${13}^{th}$ term of the given G.P. is $128$.

(iii) $\sqrt{3},3,3\sqrt{3},....729$

$a=\sqrt{3},r=\frac{t_n}{t_{n-1}}=\frac{3}{\sqrt{3}}=\sqrt{3},n=?$,

$t^n=729$

Now,

$t_n=a{r}^{n-1}$

$729=\left(\sqrt{3}\right)(r{)}^{n-1}$

Now,

$\Rightarrow 729=\left(\sqrt{3}\right)(\sqrt{3}{)}^{n-1}$

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

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

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

$\Rightarrow n=12$

Thus, the ${12}^{th}$ term of the given G.P. is $729$.

(iv) $\frac{1}{3},\frac{1}{9},\frac{1}{27}....\frac{1}{19683}$

$a=\frac{1}{3},r=\frac{t_n}{t_{n-1}}=\frac{t_2}{t_1}=\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{1}{3},t_n=\frac{1}{19683}$,

n = ?

Now,

$t_n=a{r}^{n-1}$

$\Rightarrow \frac{1}{19683}=\left(\frac{1}{3}\right){\left(\frac{1}{3}\right)}^{n-1}={\left(\frac{1}{3}\right)}^n$

$\Rightarrow {\left(\frac{1}{3}\right)}^9={\left(\frac{1}{3}\right)}^n$

$\Rightarrow n=9$

Thus $9^{th}$ term of the given G.P is $\frac{1}{19683}$.