The sum of the first ten terms of a geometric | vedclass

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(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}} $

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(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 $r = \pm \sqrt[10]{\frac{S_2}{S_1}}$.

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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