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bonufazy [111]
2 years ago
15

What is the sum of a 6-term geometric sequined if the first term is 22 and the last term is 1299078

Mathematics
1 answer:
stira [4]2 years ago
8 0

Answer:

1461460

Step-by-step explanation:

From the question given, we obtained the following data:

a = 22

Last term = T6 = 1299078

Let us find the common ratio(r)

Tn = ar^(n-1)

T6 = ar^5

1299078 = 22r^5

Divide both side by 22

r^5 = 1299078/22

r^5 = 59049

Take the fifth root of both side

r = 5√ 59049

r = 9

Now we can find the sum of the 6th term as follows

Sn = a[(r^n) — 1] / (r —1)

S6 = 22[(9^6) — 1] / 9—1

S6 = 22[(9^6) — 1] / 9—1

S6 = 22[531441 — 1] / 8

S6 = 22[531440] /8

S6 = 22 x 66430

S6 = 1461460

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Answer:

The answer is hyperbola; (x')² - (y')² - 16 = 0 ⇒ answer (a)

Step-by-step explanation:

* At first lets talk about the general form of the conic equation

- Ax² + Bxy + Cy²  + Dx + Ey + F = 0

∵ B² - 4AC < 0 , if a conic exists, it will be either a circle or an ellipse.

∵ B² - 4AC = 0 , if a conic exists, it will be a parabola.

∵ B² - 4AC > 0 , if a conic exists, it will be a hyperbola.

* Now we will study our equation:

 xy = -8

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∴ B² - 4 AC = (1)² - 4(0)(0) = 1 > 0

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∴ The graph is hyperbola

* The equation xy = -8

∵ We have term xy that means we rotated the graph about

  the origin by angle Ф

∵ Ф = π/4

∴ We rotated the x-axis and the y-axis by angle π/4

* That means the point (x' , y') it was point (x , y)

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∴ x' = xcos(π/4) - ysin(π/4) , y' = xsin(π/4) + ycos(π/4)

∴ x' = x/√2 - y/√2 = (x - y)/√2

∴ y' = x/√2 + y/√2 = (x + y)/√2

* Lets substitute x' and y' in the 1st answer

∵ (x')² - (y')² - 16 = 0

∴ (\frac{x-y}{\sqrt{2}})^{2}-(\frac{x+y}{\sqrt{2}})^{2}=

 ( \frac{x^{2}-2xy+y^{2}}{2})-(\frac{x^{2}+2xy+y^{2}}{2})-16=0

* Lets open the bracket

∴ \frac{x^{2}-2xy+y^{2}-x^{2}-2xy-y^{2}}{2}-16=0

* Lets add the like terms

∴ \frac{-4xy}{2}-16=0

* Simplify the fraction

∴ -2xy - 16 = 0

* Divide the equation by -2

∴ xy + 8 = 0

∴ xy = -8 ⇒ our equation

∴ Answer (a) is our answer

∴ The answer is hyperbola; (x')² - (y')² - 16 = 0

* Look at the graph:

- The black is the equation (x')² - (y')² - 16 = 0

- The purple is the equation xy = -8

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Now we got the smaller number, which is 141, we can also find the larger number by adding 2 to it. 141+2 = 143, 
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