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brittyarellano

  • one year ago

what is the 32nd term of the arithmetic sequence where a1=13 and a13=-59

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  1. jdoe0001
    • one year ago
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    recall \(\Large a_{\color{brown}{ n}}=a_1+({\color{brown}{ n}}-1)d ?\)

  2. brittyarellano
    • one year ago
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    yes but I'm having trouble being able to find d

  3. jdoe0001
    • one year ago
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    well, we know the 13th term is 59 and the 1st one is 13 so, we could say that \(\bf a_{\color{brown}{ 13}}={\color{blue}{ a_1}}+({\color{brown}{ 13}}-1)d\implies 59={\color{blue}{ 13}}+({\color{brown}{ 13}}-1)d\) so, to find "d", just solve for "d" :)

  4. brittyarellano
    • one year ago
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    yes but I'm having trouble being able to find d

  5. jdoe0001
    • one year ago
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    once you find "d", you can find \(\Large a_{32}\)

  6. brittyarellano
    • one year ago
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    yes but I'm having trouble being able to find d

  7. brittyarellano
    • one year ago
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    i tried doing that and what I got wasn't one of the answer choices. It gave me the options of -185,-179,-173- and -167

  8. jdoe0001
    • one year ago
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    well.. what did you get for "d"?

  9. brittyarellano
    • one year ago
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    I got -2.36

  10. jdoe0001
    • one year ago
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    hmmm actually is -59 but anyhow \(\bf a_{\color{brown}{ 13}}={\color{blue}{ a_1}}+({\color{brown}{ 13}}-1)d\implies -59={\color{blue}{ 13}}+({\color{brown}{ 13}}-1)d\)

  11. brittyarellano
    • one year ago
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    I plugged it in with -59 and I got -2.36

  12. jdoe0001
    • one year ago
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    well... hold the mayo.. .i messed up the signs again, shoot

  13. jdoe0001
    • one year ago
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    \(\bf a_{\color{brown}{ 13}}={\color{blue}{ a_1}}+({\color{brown}{ 13}}-1)d\implies -59={\color{blue}{ 13}}+({\color{brown}{ 13}}-1)d \\ \quad \\ -59-13=12d\implies -72=12d\implies \cfrac{\cancel{-72}}{\cancel{12}}=d \\ \quad \\ -6=d\)

  14. brittyarellano
    • one year ago
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    oh I see where I messed. I got -173 for the answer! thanks so much!

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