For any positive integer n, the sum of the first n positive integers equals (n(n+1))/2. what is teh sum of all the even integers between 99 and 301.
The correct answer in the OG is 20,200. The OG solves this problem by plugging 159 and 49 as n because " the sum of the even integers between 99 and 301 is the sum of the even integers from 100 through 300 or the sum of the 50th even integer through the 150 even integers. However, when solving the problem they multiply each of the equation by two: 2((150(150+1))/2) - 2((49(49+1)/2).
For the life of me i can't understand why they multiply the two formula by two. Can anyone explain?
OG Problem #157
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- Rahul@gurome
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The sum of all even integers between 99 and 301 is 100 + 102 + 104.....+300 = 2(50+51+52+...+150)
= 2*(1+2+.....+150) - 2*(1+2+...+49)
=2*150*(151)/2 - 2*49*(50)/2 = 20,200.
= 2*(1+2+.....+150) - 2*(1+2+...+49)
=2*150*(151)/2 - 2*49*(50)/2 = 20,200.
Rahul Lakhani
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I get the first sentence completely but can you explain the next two expressions?
= 2*(1+2+..+150) - 2*(1+2+.....+49) why did you multiply the two expression by 2
=2*150*(151)/2 - 2*49*(50)/2 = 20,200. and here divide by 2?
= 2*(1+2+..+150) - 2*(1+2+.....+49) why did you multiply the two expression by 2
=2*150*(151)/2 - 2*49*(50)/2 = 20,200. and here divide by 2?
- rishab1988
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Here is my approach
Avg = Sum/no of terms
Therefore, sum = Avg * no of terms
For no of terms:
First term =100 [first even integer]
Last term =300 [last even integer]
No of terms = (300-100)/2 [for the even integers are spaced 2 apart]
= (200)/2 =100
Add 1 to this for the terms we are talking about are 100 to 300 inclusive
No of terms = 100+1 = 101
Average in Any Arithmetic progression [ a sequence obtained by adding a fixed constant to the previous term] :
(First term + Last term)/2 = (100+300)/2 = 200
Sum = Avg * No of terms =101* 200= 200 (100+1) =200*100 +200 = 20000+200 = 20200
Avg = Sum/no of terms
Therefore, sum = Avg * no of terms
For no of terms:
First term =100 [first even integer]
Last term =300 [last even integer]
No of terms = (300-100)/2 [for the even integers are spaced 2 apart]
= (200)/2 =100
Add 1 to this for the terms we are talking about are 100 to 300 inclusive
No of terms = 100+1 = 101
Average in Any Arithmetic progression [ a sequence obtained by adding a fixed constant to the previous term] :
(First term + Last term)/2 = (100+300)/2 = 200
Sum = Avg * No of terms =101* 200= 200 (100+1) =200*100 +200 = 20000+200 = 20200
- rishab1988
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As to explanation for Rahul's solution in second line.
The sentence 2*[1+2+.....+150] - 2[1+2+3..+ 49] basically is doing the simpl function of subtracting the sum of all even integers from 2 to 98 from the sum of all even integers from 2 to 300.[ this is basically a tweak to fit in the formula n(n+1)/2]
Subtracting,we basically get the sum of all integers from 100 to 300.
I hope this clears your doubt
The sentence 2*[1+2+.....+150] - 2[1+2+3..+ 49] basically is doing the simpl function of subtracting the sum of all even integers from 2 to 98 from the sum of all even integers from 2 to 300.[ this is basically a tweak to fit in the formula n(n+1)/2]
Subtracting,we basically get the sum of all integers from 100 to 300.
I hope this clears your doubt