Find the sum of $n$ terms in the geometric progression $\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots$
The given $G.P.$ is $\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots$
Here, $a=\sqrt{7}$ and $r=\frac{\sqrt{21}}{7}=\sqrt{3}$
$S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$
$\Rightarrow S_{n}=\frac{\sqrt{7}\left[1-(\sqrt{3})^{n}\right]}{1-\sqrt{3}}$
$\Rightarrow S_{n}=\frac{\sqrt{7}\left[1-(\sqrt{3})^{n}\right]}{1-\sqrt{3}} \times \frac{1+\sqrt{3}}{1+\sqrt{3}}$
$\Rightarrow S_{n}=\frac{\sqrt{7}(\sqrt{3}+1)\left[1-(\sqrt{3})^{n}\right]}{1-3}$
$\Rightarrow S_{n}=\frac{-\sqrt{7}(\sqrt{3}+1)\left[1-(\sqrt{3})^{n}\right]}{2}$
$\Rightarrow \frac{\sqrt{7}(1+\sqrt{3})}{2}\left[(3)^{\frac{n}{2}}-1\right]$
$0.14189189189….$ can be expressed as a rational number
Find the sum of the following series up to n terms:
$6+.66+.666+\ldots$
If the ratio of the sum of first three terms and the sum of first six terms of a $G.P.$ be $125 : 152$, then the common ratio r is
The sum of the $3^{rd}$ and the $4^{th}$ terms of a $G.P.$ is $60$ and the product of its first three terms is $1000$. If the first term of this $G.P.$ is positive, then its $7^{th}$ term is
If ${a^2} + a{b^2} + 16{c^2} = 2(3ab + 6bc + 4ac)$, where $a,b,c$ are non-zero numbers. Then $a,b,c$ are in