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

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

Mathematics
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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
:D
but is the first 20 Positive integers . . .

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and it says the answer is 1960
  • M
i will put my money on daniel heh but not sure how he got that formula sorry
haha okay thanks :)
u got the answer?
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
10(6 + 19(10))
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
sum of an arithmetic progression (n/2) [2a + (n-1)d]
^ Yup. That's the generalized version. Cleaned up a little bit:\[n/2 (2a + (n-1)d)\]
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...
hahahaha...lol
you deserved a medal for that one :P

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