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anonymous

  • 5 years ago

Find the sum of the following: First 20 positive integers ending in 3

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  1. anonymous
    • 5 years ago
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    3, 13, etc... Rule: 3+10(n-1) = 10n-7, n ranges from 1 to 20. Sum of that = (10*20-7 +3)*10/2 = 1960/2 = 980

  2. anonymous
    • 5 years ago
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    :D

  3. anonymous
    • 5 years ago
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    but is the first 20 Positive integers . . .

  4. anonymous
    • 5 years ago
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    and it says the answer is 1960

  5. M
    • 5 years ago
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    i will put my money on daniel heh but not sure how he got that formula sorry

  6. anonymous
    • 5 years ago
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    haha okay thanks :)

  7. anonymous
    • 5 years ago
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    u got the answer?

  8. anonymous
    • 5 years ago
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    I divided by 2 twice. It should have been 196*20/2, not 196*10/2 (I second-guessed myself). Your book is right :D

  9. anonymous
    • 5 years ago
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    10(6 + 19(10))

  10. anonymous
    • 5 years ago
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    The formula is an adaptation old Gaussian equation n(n+1)/2. He did it by saying that: 1 + 2 + 3 + 4 + ... + n-1 + n + n + n-1 + ... + 4 + 3 + 2 + 1 = (n+1)n/2 What I did: 3 + 13 + 23 + ... + 183 + 193 +193 + 183 + ... + 23 + 13 + 3 =196*(20)/2 = 1960

  11. anonymous
    • 5 years ago
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    sum of an arithmetic progression (n/2) [2a + (n-1)d]

  12. anonymous
    • 5 years ago
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    ^ Yup. That's the generalized version. Cleaned up a little bit:\[n/2 (2a + (n-1)d)\]

  13. anonymous
    • 5 years ago
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    Sorry about the first answer UkaLailey. I'm going to start a thread that asks if I should go into retirement... I've missed 2 today...

  14. anonymous
    • 5 years ago
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    hahahaha...lol

  15. anonymous
    • 5 years ago
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    you deserved a medal for that one :P

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