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गुणोत्तर श्रेणी का योगफल निर्दिष्ट पदों तक ज्ञात कीजिए।
$\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots n$ पदों तक
$\frac{\sqrt{7}(1+\sqrt{3})}{2}\left[(3)^{\frac{n}{2}}-1\right]$
$\frac{\sqrt{7}(1+\sqrt{3})}{2}\left[(3)^{\frac{n}{2}}-1\right]$
$\frac{\sqrt{7}(1+\sqrt{3})}{2}\left[(3)^{\frac{n}{2}}-1\right]$
$\frac{\sqrt{7}(1+\sqrt{3})}{2}\left[(3)^{\frac{n}{2}}-1\right]$
Solution
The given $G.P.$ is $\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots$
Here, $a=\sqrt{7}$ and $r=\frac{\sqrt{21}}{7}=\sqrt{3}$
$S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$
$\Rightarrow S_{n}=\frac{\sqrt{7}\left[1-(\sqrt{3})^{n}\right]}{1-\sqrt{3}}$
$\Rightarrow S_{n}=\frac{\sqrt{7}\left[1-(\sqrt{3})^{n}\right]}{1-\sqrt{3}} \times \frac{1+\sqrt{3}}{1+\sqrt{3}}$
$\Rightarrow S_{n}=\frac{\sqrt{7}(\sqrt{3}+1)\left[1-(\sqrt{3})^{n}\right]}{1-3}$
$\Rightarrow S_{n}=\frac{-\sqrt{7}(\sqrt{3}+1)\left[1-(\sqrt{3})^{n}\right]}{2}$
$\Rightarrow \frac{\sqrt{7}(1+\sqrt{3})}{2}\left[(3)^{\frac{n}{2}}-1\right]$