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lavalava Group Title

The fourth term of an arithmetic sequence is 141, and the seventh term is 132. The first term is _____.

  • 11 months ago
  • 11 months ago

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  1. satellite73 Group Title
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    \[a, a+d,a+2d,a+3d,a+4d,...\] you got \(a+3d=141\) and \(a+6d=132\)

    • 11 months ago
  2. satellite73 Group Title
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    that means \(132-141=(a+6d)-(a+3d)=3d=-9\) and so \(d=-3\)

    • 11 months ago
  3. lavalava Group Title
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    okay so then d=4?

    • 11 months ago
  4. satellite73 Group Title
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    actually \(d=-3\)

    • 11 months ago
  5. lavalava Group Title
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    umm please explain!!

    • 11 months ago
  6. satellite73 Group Title
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    k lets go slow

    • 11 months ago
  7. lavalava Group Title
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    please and thanks

    • 11 months ago
  8. satellite73 Group Title
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    actually before we even start, since \(132<141\) is it clear that the terms are getting smaller?

    • 11 months ago
  9. satellite73 Group Title
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    in other words, "\(d\)" the "common difference" must be negative right?

    • 11 months ago
  10. lavalava Group Title
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    okay

    • 11 months ago
  11. satellite73 Group Title
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    if we call the first term \(a\) then the second term is \(a+d\) for some \(d\) and the third term is \(a+2d\), the fourth term is \(a+3d\) the fifth term is \(a+4d\) etc

    • 11 months ago
  12. satellite73 Group Title
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    in other words, you keep adding \(d\) to each term to get the next term, with the sophistication that you might be "adding" a negative number

    • 11 months ago
  13. lavalava Group Title
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    okay

    • 11 months ago
  14. satellite73 Group Title
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    The fourth term of an arithmetic sequence is 141 tells you that \(a+3d=141\)

    • 11 months ago
  15. lavalava Group Title
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    okay

    • 11 months ago
  16. satellite73 Group Title
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    you see that it is the fourth term, so it is \(a+3d\) not \(a+4d\)

    • 11 months ago
  17. lavalava Group Title
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    ohhh

    • 11 months ago
  18. satellite73 Group Title
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    and the seventh term is 132 means \[a+6d=132\]

    • 11 months ago
  19. satellite73 Group Title
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    from these two pieced of information we can solve for \(d\) and then solve for \(a\)

    • 11 months ago
  20. satellite73 Group Title
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    *pieces

    • 11 months ago
  21. lavalava Group Title
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    okay

    • 11 months ago
  22. satellite73 Group Title
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    a bit of algebra shows that \[a+6d-(a+3d)=3d\] right?

    • 11 months ago
  23. lavalava Group Title
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    ohh okay i get it..

    • 11 months ago
  24. satellite73 Group Title
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    so we see that \[3d=132-141=-9\]

    • 11 months ago
  25. lavalava Group Title
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    mhm

    • 11 months ago
  26. satellite73 Group Title
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    so far so good?

    • 11 months ago
  27. lavalava Group Title
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    and so since it is 7-4= 3 then -9/3 would be -3 right? giving us the difference

    • 11 months ago
  28. satellite73 Group Title
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    yeah \(-3\) is the difference

    • 11 months ago
  29. satellite73 Group Title
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    what you said

    • 11 months ago
  30. lavalava Group Title
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    ohhh okay!!! i get it!! :D

    • 11 months ago
  31. satellite73 Group Title
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    you are still not done though right?

    • 11 months ago
  32. satellite73 Group Title
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    your question asked "The first term is _____"

    • 11 months ago
  33. lavalava Group Title
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    hmm well pluggin in the difference and then your equation... a4=141 a3=141+3=144 a2=144+3=147 a1=147+3=150 so then the first term would be 150 right? i just did it backwards

    • 11 months ago
  34. satellite73 Group Title
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    yeah i guess so i would have said \(a+3\times (-3)=141\)or \[a-9=141\] making \(a=150\) your method means you understand what is going on, which is good but you certainly wouldn't want to use that if you had say \(a_{75}\) and wanted \(a_1\)

    • 11 months ago
  35. lavalava Group Title
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    :D yay thank you!!! :D ohh okay... ill keep in mind that equation!! thank you soo much!

    • 11 months ago
  36. satellite73 Group Title
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    yw

    • 11 months ago
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