How many terms of $G.P.$ $3,3^{2}, 3^{3}$... are needed to give the sum $120 ?$
The given $G.P.$ is $3,3^{2}, 3^{3} \ldots$
Let $n$ terms of this $G.P.$ be required to obtain in the sum as $120 .$
$S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$
Here, $a=3$ and $r=3$
$\therefore S_{n}=120=\frac{3\left(3^{n}-1\right)}{3-1}$
$\Rightarrow 120=\frac{3\left(3^{n}-1\right)}{2}$
$\Rightarrow \frac{120 \times 2}{3}=3^{n}-1$
$\Rightarrow 3^{n}-1=80$
$\Rightarrow 3^{n}=81$
$\Rightarrow 3^{n}=3^{4}$
$\therefore n=4$
Thus, four terms of the given $G.P.$ are required to obtain the sum as $120 .$
Let $S_1$ be the sum of areas of the squares whose sides are parallel to coordinate axes. Let $S_2$ be the sum of areas of the slanted squares as shown in the figure. Then, $\frac{S_1}{S_2}$ is equal to
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 $G.P.$
In a increasing geometric series, the sum of the second and the sixth term is $\frac{25}{2}$ and the product of the third and fifth term is $25 .$ Then, the sum of $4^{\text {th }}, 6^{\text {th }}$ and $8^{\text {th }}$ terms is equal to
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
The two geometric means between the number $1$ and $64$ are