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Geometric Progression

The sum of fi...

Question

The sum of first $n$ terms of an G.P. is _____ when $| r | < 1$ .

S_{n} = \frac{a_{1} (1 - r^{n})}{1 - r}

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S_{n} = \frac{a_{1} (1 + r^{n})}{1 - r}

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S_{n} = \frac{a_{1} (1 - r^{n})}{1 + r}

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S_{n} = \frac{a_{1} (1 - r^{n})}{r - 1}

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Solution

The correct option is A $S_{n} = \frac{a_{1} (1 - r^{n})}{1 - r}$
A GP can be written as:
$a, a r, a r^{2}, a r^{3} . . . . . . . . . . . . . ., a r^{n - 1}$
$sum = a + a r + a r^{2} + a r^{3} + . . . . . . . . . + a r^{n - 1}$
$sum = a (r^{n - 1} + r^{n - 2} + r^{n - 3} + r^{n - 4} + . . . . . . . . . . . + r + 1)$
We know that:
$\frac{x^{n} - 1}{x - 1} = x^{n - 1} + x^{n - 2} + x^{n - 3} + . . . . . . . . . . . . . + x + 1$
Thus $sum = a (\frac{r^{n} - 1}{r - 1})$

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