If [tex]a + ar + ar^{2} + ... = 15[/tex] and [tex]a + ar^{2} + ar^{4} + ar^{6} + ... = 9[/tex], then what is the sum of the geometric series [tex]a + ar^{3} + ar^{6} + ar^{9} + ...[/tex]?

Answers

Answer 1

Notice that

a + ar³ + ar⁶ + ar⁹ + ... = a + r³ (a + ar³ + ar⁶ + ...)

but the grouped sum on the right side is the same as the sum on the left side. This means that, if the sum we want to compute

a + ar³ + ar⁶ + ar⁹ + ...

converges to S, then

S = a + r³ S

Solve for S :

S - r³ S = a

(1 - r³) S = a

S = a/(1 - r³)

Similarly, knowing the value of the first sum, we have

a + ar + ar² + ar³ + ... = a + r (a + ar + ar² + ...)

but we know

a + ar + ar² + ... = 15

so

15 = a + 15r

and knowing the value of the second sum, we have

a + ar² + ar⁴ + ... = a + r² (a + ar² + ar⁴ + ...)

so

9 = a + 9r²

Solve these two equations for a and r :

15 = a + 15r   ⇒   a = 15 - 15r

9 = a + 9r²   ⇒   9 = 15 - 15r + 9r²   ⇒   3r² - 5r + 2 = (3r - 2) (r - 1) = 0

⇒   r = 2/3 or r = 1

We cannot have r = 1, since that would mean

a + a + a + ... = 15

since the left side is adding infinitely many copies of some constant; such a sum can never converge to a finite number. So it must be that r = 2/3, and it follows that a = 15 - 15 (2/3) = 5.

Then the sum we want has a value of

S = 5/(1 - (2/3))³ = 135/19

Answer 2

Answer:

Step-by-step explanation:

  a + ar¹ + ar² + ar³ + ... = 15       #1

r(a + ar¹ + ar² + ar³ + ...) = 15r

       ar¹ + ar² + ar³ + ... = 15r      #2

subtract #2 from #1

a = 15 - 15r

     a + ar² + ar⁴ + ar⁶ + ... = 9         #3

r²(a + ar² + ar⁴ + ar⁶ + ... )= 9r²

         ar² + ar⁴ + ar⁶ + ... = 9r²       #4

subtract #4 from #3

a = 9 - 9r²

15 - 15r = 9 - 9r²

9r² - 15r + 6 = 0

3r² - 5r + 2 = 0

r = (5 ±√(5² - 4(3)(2))) / (2(3)

r = (5 ± 1) / 6

r = 1

or

r = 4/6 = 2/3

if r = 1 then a = 15 - 15(1) = 0 and a series of zeros added together can never equal 15 so r = 2/3 must be true.

a = 15 - 15(2/3) = 5

   a + ar³ + ar⁶ + ar⁹ + ... = C           #5

r³(a + ar³ + ar⁶ + ar⁹ + ...) = Cr³

         ar³ + ar⁶ + ar⁹ + ... = Cr³          #6

subtract #6 from #5

a = C - Cr³

a = C(1 - r³)

C = a/(1 - r³)

C = 5/(1 - (2/3)³)

C = 5/(1 - 8/27)

C = 5/(19/27)

C = 5(27/19)

C = 135/19 = 7.1052631578947368....


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r u just cheating ur way through a whole test lol

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Answers

Answer:12 accorns

Step-by-step explanation:

If they each hid the same number of acorns, although the squirrel needed 4 fewer holes, the chipmunk hid 3 acorns in each hole

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According to the information provided, the chipmunk hid 3 acorns in each of the holes it dug, so x = 3. The squirrel hid 4 acorns in each of the holes it dug, so y = 4.

Both the chipmunk and squirrel hid the same number of acorns, so cx = sy.

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Now, we can use the information to set up a system of equations:

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Expand the equation:

c² - 4c = 3c - 12

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c² - 4c - 3c + 12 = 0

Combine like terms:

c² - 7c + 12 = 0

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Answer:

85 ft^3 portland cement255 f^3 sand425 ft^3 crushed stone

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From the way the problem is given, there is no solution.

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