If the $4^{\text {th }}, 10^{\text {th }}$ and $16^{\text {th }}$ terms of a $G.P.$ are $x, y$ and $z,$ respectively. Prove that $x,$ $y, z$ are in $G.P.$
Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
According to the given condition,
$a_{4}=a r^{3}=x$ .......$(1)$
$a_{10}=a r^{9}=y$ .......$(2)$
$a_{16}=a r^{15}=z$ .......$(3)$
Dividing $(2)$ by $(1),$ we obtain
$\frac{y}{x}=\frac{a r^{9}}{a r^{3}} \Rightarrow \frac{y}{x}=r^{6}$
Dividing $(3)$ by $(2),$ we obtain
$\frac{z}{y}=\frac{a r^{15}}{a r^{9}} \Rightarrow \frac{z}{y}=r^{6}$
$\therefore \frac{y}{x}=\frac{z}{y}$
Thus, $x, y, z$ are in $G.P.$
Let $P(x)=1+x+x^2+x^3+x^4+x^5$. What is the remainder when $P\left(x^{12}\right)$ is divided by $P(x)$ ?
Which term of the following sequences:
$\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$ is $\frac{1}{19683} ?$
The G.M. of the numbers $3,\,{3^2},\,{3^3},\,......,\,{3^n}$ is
The first term of a $G.P.$ is $1 .$ The sum of the third term and fifth term is $90 .$ Find the common ratio of $G.P.$
Let $\alpha$ and $\beta$ be the roots of $x^{2}-3 x+p=0$ and $\gamma$ and $\delta$ be the roots of $x^{2}-6 x+q=0 .$ If $\alpha$ $\beta, \gamma, \delta$ form a geometric progression. Then ratio $(2 q+p):(2 q-p)$ is