How do you write a geometric series for which r=1/2 and n=4?

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Answer 1 · 2016-12-05T08:49:36

$a_{1}, \frac{a_{1}}{2}, \frac{a_{1}}{4}, \frac{a_{1}}{8}$ $a_1, a_1/2, a_1/4, a_1/8$ Where $a_{1}$ $a_1$ is the first term in the series

In general the $n^{t h}$ $n^(th)$ term of a geometric sequence is given by:

$a_{n} = a_{n-1} \cdot r$ $a_n = a_"n-1"*r$ Where $r$ $r$ is the common ratio and $a_{1}$ $a_1$ is the first term.

In this example, $r = \frac{1}{2}$ $r=1/2$ and $n = 4$ $n=4$

$a_{2} = a_{1} \cdot \frac{1}{2}$ $a_2 = a_1* 1/2$

$a_{3} = a_{2} \cdot \frac{1}{2} = a_{1} \cdot \frac{1}{4}$ $a_3 = a_2*1/2 = a_1*1/4$

$a_{4} = a_{3} \cdot \frac{1}{2} = a_{1} \cdot \frac{1}{8}$ $a_4 = a_3*1/2 = a_1*1/8$

In general, $a_{n} = \frac{a_{1}}{2^{n - 1}}$ $a_n = a_1/2^(n-1)$

Hence: the series requested is:

$a_{1}, \frac{a_{1}}{2}, \frac{a_{1}}{4}, \frac{a_{1}}{8}$ $a_1, a_1/2, a_1/4, a_1/8$ Where $a_{1}$ $a_1$ is the first term in the series