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anonymous
 5 years ago
Find the sum of the following: First 20 positive integers ending in 3
anonymous
 5 years ago
Find the sum of the following: First 20 positive integers ending in 3

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.03, 13, etc... Rule: 3+10(n1) = 10n7, n ranges from 1 to 20. Sum of that = (10*207 +3)*10/2 = 1960/2 = 980

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but is the first 20 Positive integers . . .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and it says the answer is 1960

M
 5 years ago
Best ResponseYou've already chosen the best response.0i will put my money on daniel heh but not sure how he got that formula sorry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I divided by 2 twice. It should have been 196*20/2, not 196*10/2 (I secondguessed myself). Your book is right :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The formula is an adaptation old Gaussian equation n(n+1)/2. He did it by saying that: 1 + 2 + 3 + 4 + ... + n1 + n + n + n1 + ... + 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sum of an arithmetic progression (n/2) [2a + (n1)d]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0^ Yup. That's the generalized version. Cleaned up a little bit:\[n/2 (2a + (n1)d)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry 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...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you deserved a medal for that one :P
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