(B) Let the first term be $a$ and the common ratio be $r$.$S_1 = a + ar + ar^2 + \dots + ar^9 = a\frac{r^{10}-1}{r-1}$.$S_2 = ar^{10} + ar^{11} + \dot

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$ \pm \sqrt[10]{\frac{S_2}{S_1}} $

(B) Let the first term be $a$ and the common ratio be $r$.$S_1 = a + ar + ar^2 + \dots + ar^9 = a\frac{r^{10}-1}{r-1}$.$S_2 = ar^{10} + ar^{11} + \dots + ar^{19} = ar^{10}(1 + r + r^2 + \dots + r^9) = ar^{10}\frac{r^{10}-1}{r-1}$.Dividing $S_2$ by $S_1$,we get:$\frac{S_2}{S_1} = \frac{ar^{10}(\frac{r^{10}-1}{r-1})}{a(\frac{r^{10}-1}{r-1})} = r^{10}$.Therefore,$r^{10} = \frac{S_2}{S_1}$,which implies $r = \pm \sqrt[10]{\frac{S_2}{S_1}}$.

Class 11 Mathematics Sequences and Series Geometric progression

Medium

The sum of the first ten terms of a geometric progression is $S_1$ and the sum of the next ten terms ($11^{th}$ to $20^{th}$) is $S_2$. What is the common ratio?

A
$ \pm \sqrt[10]{\frac{S_1}{S_2}} $
B
$ \pm \sqrt[10]{\frac{S_2}{S_1}} $
C
$ \pm 10\sqrt{\frac{S_2}{S_1}} $
D
$ \pm \sqrt{\frac{S_2}{S_1}} $

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Class 11

Mathematics Questions

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Sequences and Series

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Geometric progression

If the sum of $n$ terms of a $G.P.$ is $255$,the $n^{th}$ term is $128$,and the common ratio is $2$,then the first term will be:

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Let $a_1, a_2, a_3, \ldots$ be a $G.P.$ of increasing positive numbers. Let the sum of its $6^{\text{th}}$ and $8^{\text{th}}$ terms be $2$ and the product of its $3^{\text{rd}}$ and $5^{\text{th}}$ terms be $\frac{1}{9}$. Then $6(a_2 + a_4)(a_4 + a_6)$ is equal to

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If the $n^{th}$ term of the geometric progression $5, - \frac{5}{2}, \frac{5}{4}, - \frac{5}{8}, \dots$ is $\frac{5}{1024}$,then the value of $n$ is:

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If the sum of the first $6$ terms is $9$ times the sum of the first $3$ terms of the same $G.P.$,then the common ratio of the series will be

Easy

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If $a = r + r^2 + r^3 + \dots + \infty$,then the value of $r$ is .......

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The sum of the first ten terms of a geometric progression is $S_1$ and the sum of the next ten terms ($11^{th}$ to $20^{th}$) is $S_2$. What is the common ratio?

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