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maria-17

  • 2 years ago

Please help What is the sum of the arithmetic sequence 134, 122, 110 …, if there are 32 terms?

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  1. tpaulus
    • 2 years ago
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    Find the general form, then use that to find the last term. then use the sum of an arithmetic sequence equation to find the sum.\[s = n/2 * (n1 + n32)\]

  2. campbell_st
    • 2 years ago
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    well the sum is simply \[S_{n} = \frac{n}{2}[2a + (n -1)d]\] a = 1st term, n = number of terms and d = common difference in your question you know n = 32, the 1st term is 134 find the common difference. ..d and then substitute and evaluate. the value of d will be negative

  3. maria-17
    • 2 years ago
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    –1,664 –1,632 –1,600 –1,568

  4. maria-17
    • 2 years ago
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    Those are the answers.^

  5. tcarroll010
    • 2 years ago
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    S = [134 - (0)(12)] + [134 - (1)(12)] + [134 - (2)(12)] + . . . + [134 - (31)(12)] S = (134)(32) - (12)(1 + 2 + 3 + . . . + 31) S = (134)(32) - (12)(31)(32)/2

  6. tcarroll010
    • 2 years ago
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    All good now, @maria-17 ?

  7. maria-17
    • 2 years ago
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    THanksss.

  8. tcarroll010
    • 2 years ago
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    uw!

  9. tcarroll010
    • 2 years ago
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    Good luck to you in all of your studies and thx for the recognition! @maria-17

  10. maria-17
    • 2 years ago
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    Can you help me with one more?

  11. tcarroll010
    • 2 years ago
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    Sure, why not? Best though to close this one and just start another post.

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