anonymous
  • anonymous
Find the sum of the following: First 20 positive integers ending in 3
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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
anonymous
  • anonymous
:D
anonymous
  • anonymous
but is the first 20 Positive integers . . .

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.