In a $G.P.,$ the $3^{rd}$ term is $24$ and the $6^{\text {th }}$ term is $192 .$ Find the $10^{\text {th }}$ term.
Here, $a_{3}=a r^{2}=24$ ........$(1)$
and $a_{6}=a r^{5}=192$ ...........$(2)$
Dividing $(2)$ by $(1),$ we get $r=2 .$ Substituting $r=2$ in $(1),$ we get $a=6$
Hence $\quad a_{10}=6(2)^{9}=3072$
The sum of first three terms of a $G.P.$ is $16$ and the sum of the next three terms is
$128.$ Determine the first term, the common ratio and the sum to $n$ terms of the $G.P.$
If the ${5^{th}}$ term of a $G.P.$ is $\frac{1}{3}$ and ${9^{th}}$ term is $\frac{{16}}{{243}}$, then the ${4^{th}}$ term will be
If three successive terms of a$G.P.$ with common ratio $r(r>1)$ are the lengths of the sides of a triangle and $[\mathrm{r}]$ denotes the greatest integer less than or equal to $r$, then $3[r]+[-r]$ is equal to :
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 $P^{2}=(a b)^{n}$
Insert three numbers between $1$ and $256$ so that the resulting sequence is a $G.P.$