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anonymous

  • 4 years ago

find sum of the 1st 20 terms in the arithmetic series 5,8,11,14

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  1. amistre64
    • 4 years ago
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    what do we add to get to each next term?

  2. amistre64
    • 4 years ago
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    called a common difference

  3. anonymous
    • 4 years ago
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    3

  4. amistre64
    • 4 years ago
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    3 is right :) so, Sn = n(first+last)/2 if i recall it right

  5. amistre64
    • 4 years ago
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    to determine the 20th term is a20 = 5+3(19) = 62 62+5 = 67*20/2 = 670 ... if my math was good

  6. anonymous
    • 4 years ago
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    Oh so it's 670? Thank you so so so much! wow thank you.

  7. amistre64
    • 4 years ago
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    i hope so, id dbl chk with the wolf: sum (5+3(n-1)) from 1 to 20

  8. amistre64
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=sum+%285%2B3%28n-1%29%29+from+1+to+20

  9. amistre64
    • 4 years ago
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    670 :)

  10. anonymous
    • 4 years ago
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    Can you help me with one more? Find the 7th term in the geometric sequence 4,20,11

  11. amistre64
    • 4 years ago
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    20/4 = our common ratio an = a1 * r^(n-1) a7 = 4 * (20/5)^6

  12. amistre64
    • 4 years ago
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    your 11 is bad there, is that a typo?

  13. anonymous
    • 4 years ago
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    sorry i meant 100

  14. anonymous
    • 4 years ago
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    yes :/

  15. amistre64
    • 4 years ago
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    whew!! lol, then yeah; 4*5^6 should do it

  16. anonymous
    • 4 years ago
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    so a^7 = 4*5^6?

  17. anonymous
    • 4 years ago
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    dd you get 250?

  18. amistre64
    • 4 years ago
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    i think its bigger than that; 5^4 = 25^2 = 625 alone

  19. amistre64
    • 4 years ago
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    625*20 is what id end up with

  20. amistre64
    • 4 years ago
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    12500 perhaps? hard to do in the head :)

  21. amistre64
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=4*%285%5E6%29 almost had it :)

  22. amistre64
    • 4 years ago
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    i left off the last *5 .... 12500*5=62500

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