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anonymous

  • one year ago

Find the sum of the first 12 terms of the sequence. Show all work for full credit. 1, -4, -9, -14, . . . please this is my last question and i cant seem to get it right. if possible can someone work it out for me?

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  1. campbell_st
    • one year ago
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    what do you know about the question..?

  2. anonymous
    • one year ago
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    i believe its an arithmetic sequence?

  3. campbell_st
    • one year ago
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    good start, what's the common difference..?

  4. anonymous
    • one year ago
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    ok that where i confuse myself

  5. campbell_st
    • one year ago
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    well what is the sequence changing by... common difference = 2nd term - 1st term check with 3rd term - 2nd term.... is it the same

  6. anonymous
    • one year ago
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    -5

  7. campbell_st
    • one year ago
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    great... and the 1st term a = 1 so for the sum of the sequence you can use the formula \[S_{n} = \frac{n}{2}[2a + (n - 1) \times d]\] you are given a = 1, you found d = -5 and have been told n = 12 so substitute them and calculate the sum

  8. anonymous
    • one year ago
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    i get the plugging in but what does d equal? all the rest i plugged in

  9. campbell_st
    • one year ago
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    d is the common difference d = -5

  10. anonymous
    • one year ago
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    i see ok 1 min

  11. anonymous
    • one year ago
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    is it 6/-53

  12. anonymous
    • one year ago
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    that's what i got but I not entirely sure.

  13. campbell_st
    • one year ago
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    you are close the calculation becomes \[S_{12} = \frac{12}{2} \times (-53)\]

  14. campbell_st
    • one year ago
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    so then \[S_{12} = 6 \times (-53)\]

  15. anonymous
    • one year ago
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    so that is the final form?

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