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Find the missing terms in each geometric sequence.

1. 256, ___, ___, -32, -64
2. 27, 9, ___, ___, 1/3
3. 5x^2, ___, 5x^6, 5x^8, ___, ...

Respuesta :

devishri1977

Answer:

1) 256, -128, 64, -32

2) 27, 9, 3, 1, 1/3

3) 5x², 5x⁴, 5x⁶, 5x⁸, 5x¹⁰,

Step-by-step explanation:

1. In this geometric series each term is obtained by multiplying the previous term by (-1/2) or dividing by (-2).

256 ÷ (-2) = - 128

-128 ÷ (-2) = 64

64 ÷ (-2) = (-32)

The geometric series is:

     256, -128, 64, -32...

We can find the common ratio by dividing the second term by first term.

2) 27, 9, ___, _________, 1/3

       [tex]\sf \boxed{\bf common \ ratio= \dfrac{second \ term}{first \ term}}[/tex]

                                 [tex]\sf = \dfrac{9}{27}\\\\ = \dfrac{1}{3}[/tex]

  [tex]\sf 9*\dfrac{1}{3}=3\\\\3*\dfrac{1}{3}=1[/tex]

The geometric series is:

  27, 9, 3, 1, 1/3....

3) 5x², _____, 5x⁶, 5x⁸, _____, ....

Here, we can take 3rd term and 4th term to find the common ratio.

  [tex]\sf common \ ratio = \dfrac{5x^8}{5x^6}\\\\[/tex]

                       [tex]\s = x^{8-6}\\\\=x^2[/tex]

5x² * x² = 5x⁴

8x⁸ * x² = 8x¹⁰

The geometric serious is:

5x², 5x⁴, 5x⁶, 5x⁸, 5x¹⁰, ...

fieryanswererft

Answer's:

[tex]\sf 1.) \ 256, -128, 64, -32,..\\ \\ 2.) \ 27, 9, 3, 1, 1/3,... \\ \\3.) \ 5x^2, 5x^4, 5x^6, 5x^8,5x^{10} , ...[/tex]

[tex]\sf Geometric \ formula: ar^{n-1}[/tex]

  • where 'a' is first term, 'r' is common ratio.

1)

Find the common ratio:

  • next term ÷ previous term
  • 32 ÷ -64 = -1/2

Equation: [tex]256(-\frac{1}{2} )^{n-1}[/tex]

2nd term: [tex]256(-\frac{1}{2} )^{2-1}=-128[/tex]

3rd term: [tex]256(-\frac{1}{2} )^{3-1}=64[/tex]

2)

Find common ratio:

  • next term ÷ previous term
  • 9 ÷ 27 = 1/3

Equation: [tex]27(\frac{1}{3} )^{n-1}[/tex]

3rd term: [tex]27(\frac{1}{3} )^{3-1}=3[/tex]

4th term: [tex]27(\frac{1}{3} )^{4-1}=1[/tex]

3)

Find the common ratio:

  • next term ÷ previous term
  • 5x^8 ÷ 5x^6 = x^2

Equation: [tex]5x^2 (x^2)^{n-1}[/tex]

2nd term: [tex]5x^2 (x^2)^{2-1} = 5x^2 (x^2) = 5x^{4}[/tex]

5ht term: [tex]5x^2 (x^2)^{5-1} = 5x^2 (x^8) = 5x^{10}[/tex]