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3 years ago

Write a sequence that has four geometric means between 31 and –23,540,625.

Mathematics

2 answers:

tino4ka555 [31]3 years ago

7 0

Answer:

$a_n=31(-15)^{n-1}$

31, -465, 6975, -104,625, 1,569,375, -23,540,625

Step-by-step explanation:

The formula for a geometric sequence is:

$a_n=a_1(r)^{n-1}$

The formula for a geometric sequence is:

Where

r is the common ratio

$a_1$ is the first term

$a_n$ is the nth term

In this case

$a_1=31$

$a_6=-23,540,625$

So:

$-23,540,625=31(r)^{6-1}$

Now we solve for r

$-23,540,625=31(r)^{5}$

$-\frac{23,540,625}{31}=r^{5}\\\\r=\sqrt[5]{-\frac{23,540,625}{31}}\\\\r=-15$

Then the four geometric means are

$a_2=31(-15)^{2-1}=-465$

$a_3=31(-15)^{3-1}=6975$

$a_4=31(-15)^{4-1}=-104,625$

$a_5=31(-15)^{5-1}=1,569,375$

31, -465, 6975, -104,625, 1,569,375, -23,540,625

aleksandr82 [10.1K]3 years ago

7 0

Answer:

31, -465, 6975, -104,625, 1,569,375, -23,540,625

Step-by-step explanation:

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The above question can be written like

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Step-by-step explanation:

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<img src="https://tex.z-dn.net/?f=%5Cboxed%7B%5Cblue%7B%5Cmathscr%7BHello%5C%3ABrainliacs%7D%7D%7D" id="TexFormula1" title="\box

andrew-mc [135]

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$CO(g) + H_2 O(g) = CO_2(g) + H_2(g)$

Initial concentration:

0.1M , 0.1M , 0 , 0

Let 'x' mole per litre of each of theproduct be formed.

At equilibrium:

(0.1 – x)M , (0.1 – x)M , xM , xM

where x is the amount of Carbon dioxide and Hydrogen, at equilibrium.

Hence, equilibrium constant can be written as,

$K_c = \frac{x²}{(0.1 – x)²} = 4.24$

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→ 3.24x² - 0.848x + 0.0424 = 0

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(for quadratic equation ax² + bx+c=0)

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$= > x = \frac{ - ( - 0.848 \: ± \: \sqrt{( - 0.848)^{2} - 4(3.24)(0.0424) } }{2 \times 3.24}$

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$x_1 = \frac{0.848 - 0.4118}{6.48} = 0.067$

$x_2 = \frac{0.848 + 0.4118}{6.48} = 0.194$

Here, the value 0.194 should be neglected because it will give concentration of the reactant which is more than initial concentration.

∴ The equilibrium concentrations are :-

$[CO_2] [H_2] = x = 0.067M$

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