Which term of the following sequences:
$\sqrt{3}, 3,3 \sqrt{3}, \ldots$ is $729 ?$
The given sequence is $\sqrt{3}, 3,3 \sqrt{3,}, \ldots \ldots$
$a=\sqrt{3}$ and $r=\frac{3}{\sqrt{3}}=\sqrt{3}$
Let the $n^{\text {th }}$ term of the given sequence be $729 .$
$a_{n}=a r^{n-1}$
$\therefore a r^{n-1}=729$
$\Rightarrow(\sqrt{3})(\sqrt{3})^{n-1}=729$
$\Rightarrow(3)^{1 / 2}(3)^{\frac{n-1}{2}}=(3)^{6}$
$\Rightarrow(3)^{\frac{1}{2}+\frac{n-1}{2}}=(3)^{6}$
$\therefore \frac{1}{2}+\frac{n-1}{2}=6$
$\Rightarrow \frac{1+n-1}{2}=6$
$\Rightarrow n=12$
Thus, the $12^{\text {th }}$ term of the given sequence is $729 .$
If the first and the $n^{\text {th }}$ term of a $G.P.$ are $a$ and $b$, respectively, and if $P$ is the product of $n$ terms, prove that $P^{2}=(a b)^{n}$
Find four numbers forming a geometric progression in which the third term is greater than the first term by $9,$ and the second term is greater than the $4^{\text {th }}$ by $18 .$
If in a geometric progression $\left\{ {{a_n}} \right\},\;{a_1} = 3,\;{a_n} = 96$ and ${S_n} = 189$ then the value of $n$ is
$0.\mathop {423}\limits^{\,\,\,\,\, \bullet \, \bullet \,} = $
If $a,\,b,\,c$ are in $A.P.$ and ${a^2},\,{b^2},{c^2}$ are in $H.P.$, then