Find the sum of n terms in the geometric prog | vedclass

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Question

(A) The given $G.P.$ is $\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots$Here,the first term $a = \sqrt{7}$ and the common ratio $r = \frac{\sqrt{21}}{\sqrt{7

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$\frac{\sqrt{7}(\sqrt{3}+1)}{2}\left[(3)^{\frac{n}{2}}-1\right]$

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(A) The given $G.P.$ is $\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots$Here,the first term $a = \sqrt{7}$ and the common ratio $r = \frac{\sqrt{21}}{\sqrt{7}} = \sqrt{3}$.The sum of $n$ terms of a $G.P.$ is given by $S_{n} = \frac{a(r^{n}-1)}{r-1}$.Substituting the values,we get $S_{n} = \frac{\sqrt{7}((\sqrt{3})^{n}-1)}{\sqrt{3}-1}$.To rationalize the denominator,multiply the numerator and denominator by $(\sqrt{3}+1)$:$S_{n} = \frac{\sqrt{7}((\sqrt{3})^{n}-1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$$S_{n} = \frac{\sqrt{7}(\sqrt{3}+1)(3^{\frac{n}{2}}-1)}{3-1}$$S_{n} = \frac{\sqrt{7}(\sqrt{3}+1)}{2}\left[(3)^{\frac{n}{2}}-1\right]$.

Find the sum of $n$ terms in the geometric progression $\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots$

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