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    Ayen Dueños Mod 7 Activity 6 Think Think and Lost

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    steph curry
    this is nothing just for fun okay I needed to stop talking because I don't know what I'm typing I'm her coeruieier the eqtuiaion is not that ad is just wrong

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    Ayen Dueños Mod 7 Activity 6 Think Think and Lost

    Uploaded by

    steph curry
    23 views2 pages
    this is nothing just for fun okay I needed to stop talking because I don't know what I'm typing I'm her coeruieier the eqtuiaion is not that ad is just wrong

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    this is nothing just for fun okay I needed to stop talking because I don't know what I'm typing I'm her coeruieier the eqtuiaion is not that ad is just wrong

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    23 views2 pages

    Ayen Dueños Mod 7 Activity 6 Think Think and Lost

    Uploaded by

    steph curry
    this is nothing just for fun okay I needed to stop talking because I don't know what I'm typing I'm her coeruieier the eqtuiaion is not that ad is just wrong

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 2

    Name: Ayen Dueños

    Section: Biotechnology

    Directions: SHOW YOUR SOLUTIONS. BOX YOUR FINAL ANSWER.

    1 1
    1. −4, −1 , 1, 3 , 6, …
    2 2

    a. 𝒂𝟏 = −𝟒

    𝑎2 − 𝑎1 = 𝑎4 − 𝑎3 1
    𝑎20 = −4 + (20 − 1)2
    1 1 2
    −1 − (−4) = 3 − 1 1
    2 2 𝑎20 = −4 + (19)2
    2
    1 1
    2 =2 𝑎20 = −4 + 47.5
    2 2
    𝒂𝟐𝟎 = 𝟒𝟑. 𝟓
    𝟏
    𝒅=𝟐
    𝟐
    𝑛
    𝑏. 𝑎𝑛 + (𝑛 − 1)𝑑 𝑐. 𝑆𝑛 = [2𝑎1 + (𝑛 − 1)𝑑]
    1 2
    41 = −4 + (𝑛 − 1)2 84
    2 𝑆84 = [2(1) + (84 − 1)2.5]
    1 1 2
    41 = −4 + 2 𝑛 − 2
    2 2 𝑆84 = 42 [2 + (83)2.5]
    1 1
    41 = 2 𝑛 − 6 𝑆84 = 42 [2 + (207.5)]
    2 2
    1 1 𝑆84 = 42 (209.5)
    41 + 6 = 2 𝑛
    2 2 𝑆84 = 8799
    1
    47.5 = 2 𝑛 8799 is the sum of the
    2
    41 is in the 19th term first 84 positive integers
    𝑛 = 19

    32
    2. 24, 16, ,…
    3
    𝑎1 (𝑟 𝑛 − 1)
    𝑎2 𝑎3 𝑏. 𝑎𝑛 = 𝑎1 𝑟 𝑛−1 𝑐. 𝑆𝑛 =
    𝑟−1 𝑑. 𝑎𝑛 = 𝑎1 𝑟 𝑛−1
    𝑎. =
    𝑎1 𝑎2 2 2
    6
    𝑎6 = 24( )6−1 24 ቆ൬ ൰ − 1ቇ 1024 2
    3 3 = 24( )𝑛−1
    32 𝑆6 = 729 3
    16 3 32 2
    = 𝑎6 = 24( ) −1
    243 3 6561
    24 16 𝑖𝑛 2 ∙ 𝑖𝑛 ൬ ൰
    64
    24 ൬( ) − 1൰
    128
    𝟐𝟓𝟔 729 𝑛=
    𝟐 𝟐 𝒂𝟔 = 𝑆6 = 𝑖𝑛(2) − 𝑖𝑛3
    = 𝟖𝟏 2
    𝟑 𝟑 −1
    3
    −21.893 𝒏=𝟖
    𝑆6 =
    −0.3333

    𝑺𝟔 = 𝟔𝟓. 𝟕

    1 1 1 The given series don’t have a sum because the formula in


    3. , , ,…
    128 64 32 getting the infinite geometric series is being applied only
    when │r│< 1. In this series, the ratio is equal to 2, which is
    greater than 1.

    𝑎1
    1−𝑟 (1 − 𝑟 𝑛 )
    4. 5. 𝑆6 = 𝑎1
    1 1−𝑟
    1 − (− ) 1 6
    3 2 1 − ൬2൰
    81 𝑆6 =
    3 1
    = 2
    1
    1 1
    3 2 1 −
    64
    𝑆6 =
    81 3 1
    = 2
    4 63
    3 2 64
    𝑆6 =
    3 1
    81 3 243 2
    = ∗ = 21
    21 2 21
    1 4 4 32 𝑆6 = ∗ =
    𝑆6 = 32 1 16
    1
    𝟑 2
    = 𝟔𝟎 𝟓 1 5/6 or 1.3125 is the sum of
    𝟒 𝑺𝟔 = 𝟏
    𝟏𝟔 the first 6 terms.

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