Write the recursive formula for the geometric sequence. a1 = -2 a2 = 8 a3 = -32

2 answers:

5 0

$a_1 = -2,\ a_2 = 8,\ a_3 = -32\\\\a_2=a_1\cdot(-4)\ \ \ and\ \ \ a_3=a_2\cdot(-4)\\\\ the\ recursive\ formula:\\\\ \left \{ {\big{a_1=-2\ \ \ \ \ \ \ \ \ \ \ } \atop \big{a_n=a_{n-1}\cdot(-4)}} \right.$

FrozenT [24]2 years ago

4 0

$a_1 = -2,\ \ a_2 = 8,\ \ a_3 = -32\\\\r=\frac{a_{2}}{a_{1}}=\frac{8}{-2}=-4 \\\\The \ recursive \ formula \ for \ a \ geometric \ sequence \ is \ written \ in \ the \ form:\\\\ a_n=a_{n-1} \cdot r\\ \\ a_n=a_{n-1} \cdot (-4 )$

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Diano4ka-milaya [45]

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Third, multiply both sides by x - 5. / Your problem should look like: 3x - 5 = 0
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Fifth, divide both sides by 3. / Your problem should look like: x = $\frac{5}{3}$
Sixth, from the values of x above, we have these 3 intervals to test:
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Seventh, pick a test point for each interval.

1. For the interval x < $\frac{5}{3}$ :
Let's pick x - 0. Then, $\frac{3x0-5}{0-5}$ > 0
After simplifying, we get 1 > 0 which is true.
Keep this interval.

2. For the interval $\frac{5}{3}$ < x < 5:
Let's pick x = 2. Then, $\frac{3x2-5}{2-5}$ > 0
After simplifying, we get -0.3333 > 0, which is false.
Drop this interval.

3. For the interval x > 5:
Let's pick x = 6. Then, $\frac{3x6-5}{6-5}$ > 0
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Answer: x < $\frac{5}{3}$ and x > 5

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3 years ago

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Let's do this step by step:
Subtract both sides by 7
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