What is the 7th term of the geometric sequence where a1 = -625 and a2 = 125?

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What is the 7th term of the geometric sequence where a1 = -625 and a2 = 125?

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Answer 1 · 2015-07-17T23:36:15

The seventh term of the sequence is $- \frac{1}{25}$ $color(red)(-1/25)$ .

Explanation:

We first find the common ratio $r$ $r$ by dividing a term by its preceding term.

$r = \frac{a_{2}}{a_{1}} = \frac{125}{- 625}$ $r = a_2/a_1 = 125/(-625)$

$r = - \frac{1}{5}$ $r = -1/5$

Now we use the $n^{th}$ $n^"th"$ term rule:

$t_{n} = a r^{n - 1}$ $t_n = ar^(n-1)$ , where $a$ $a$ is the first term and $r$ $r$ is the common ratio

$t_{7} = - 625 {(- \frac{1}{5})}^{7 - 1} = - 625 {(- \frac{1}{5})}^{6} = - (5^{4}) \frac{{(- 1)}^{6}}{5^{6}} = - \frac{5^{4} \times 1}{5^{6}} = - \frac{5^{4}}{5^{6}} = - \frac{1}{5^{2}}$ $t_7 = -625(-1/5)^(7-1) = -625(-1/5)^6 = -(5^4)(-1)^6/5^6 = -(5^4 × 1)/5^6 = -5^4/5^6 = -1/5^2$

$t_{7} = - \frac{1}{25}$ $t_7 = -1/25$

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What is the 7th term of the geometric sequence where a1 = -625 and a2 = 125?

1 Answer

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