What is the sum of G.P. $1 + p + p^{2} + p^{3} + p^{4} + p^{5}, (p < 1)$ ?

S_{6} = \frac{1 - p^{6}}{1 - p}

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S_{5} = \frac{1 - p^{5}}{1 - p}

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S_{4} = \frac{1 - p^{4}}{1 - p}

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S_{6} = \frac{1 - p^{6}}{1 + p}

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Solution

The correct option is A $S_{6} = \frac{1 - p^{6}}{1 - p}$
Given sequence is $1 + p + p^{2} + p^{3} + p^{4} + p^{5}$
To find the sum of the first $S_{n}$ terms of a geometric sequence using the formula
Here $a = 1, r = p, n = 6$
We know $S_{n} = \frac{a_{1} (1 - r^{n})}{1 - r}$

$\Rightarrow S_{6} = \frac{1 (1 - (p)^{6})}{1 - p}$

$\Rightarrow S_{6} = \frac{1 - p^{6}}{1 - p}$

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