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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots$

Similar Questions

The number of bacteria in a certain culture doubles every hour. If there were $30$ bacteria present in the culture originally,how many bacteria will be present at the end of $2^{\text{nd}}$ hour,$4^{\text{th}}$ hour and $n^{\text{th}}$ hour?

In a $G.P.$,the $3^{rd}$ term is $24$ and the $6^{th}$ term is $192$. Find the $10^{th}$ term.

The $5^{\text{th}}$,$8^{\text{th}}$,and $11^{\text{th}}$ terms of a $G.P.$ are $p, q$,and $s$,respectively. Show that $q^{2} = ps$.

Find the $20^{\text{th}}$ and $n^{\text{th}}$ terms of the $G.P.$ $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots$

If in a geometric progression $\{a_n\}$,$a_1 = 3$,$a_n = 96$ and $S_n = 189$,then the value of $n$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module