Question 157 from OG 12

[tomintampa]

What is the sum of all EVEN integers between 99-301?

The answer is 20,200 and I do not understand the official explanation. Please help.

[Daan]

The sum of all even integers between 99 and 301,
is the sum of all even integers from 100 to 300, inclusive.

From 100 to 300, there 300-100 + 1= 201 odd and even numbers
you have to add the one, because when substracting 100 from 300, you incorrectly do not count 100.

We can now rearrange the list from 100-300 in a more informative way:
100 + 300 = 400
101 + 299 = 400
102 + 298 = 400
103 + 297 = 400
...
etc.

This will be possible a total of 100 times, as with 201 numbers we can make 100 pairs
and then we will be left with the number 200 (the exact middle of 100 and 300).

Half of those 100 x 400, will be the result of adding even numbers.
So to get the total value of all even numbers:

1/2 x [100 x 400] + 200 (the middle number is even) =
50 x 400 + 200 = 20.000 + 200 = 20.200

-- --

So the question is quite easy once you:
1. Figure out how to easily add 100 to 300 inclusive
2. realize half the sum of all the pairs + middle number is the sum of all even numbers

[seahawk]

100+102+ .. 300
numbers in series (300-100)/2 + 1 = 101

sum = (100+300)/2 * 101 = 20200

[panacea6565]

Originally Posted by seahawk

100+102+ .. 300
numbers in series (300-100)/2 + 1 = 101

sum = (100+300)/2 * 101 = 20200

good explanation seahawk

[tomintampa]

yes, thank you

[clock60]

the question like that is on arithmetic series you must find the sum of even number between 99 and 301. the first even number is 100 and the last is 300. af first how many numbers are here between? formula is
An=A1+d(n-1). An=300.A1=100. 300=100+2(n-1) from this n=101.
the formula for the sum of series Sn=(A1+An)/2*n.
S101=(100+300)/2*101=20200.

[Elisa Pandey]

Originally Posted by Daan

The sum of all even integers between 99 and 301,
is the sum of all even integers from 100 to 300, inclusive.

From 100 to 300, there 300-100 + 1= 201 odd and even numbers
you have to add the one, because when substracting 100 from 300, you incorrectly do not count 100.

We can now rearrange the list from 100-300 in a more informative way:
100 + 300 = 400
101 + 299 = 400
102 + 298 = 400
103 + 297 = 400
...
etc.

This will be possible a total of 100 times, as with 201 numbers we can make 100 pairs
and then we will be left with the number 200 (the exact middle of 100 and 300).

Half of those 100 x 400, will be the result of adding even numbers.
So to get the total value of all even numbers:

1/2 x [100 x 400] + 200 (the middle number is even) =
50 x 400 + 200 = 20.000 + 200 = 20.200

-- --

So the question is quite easy once you:
1. Figure out how to easily add 100 to 300 inclusive
2. realize half the sum of all the pairs + middle number is the sum of all even numbers

[sandy_online]

sum of A.P = n/2[2*a+(n-1)*d]
here, series is 100,102,....,300
300 = 100+(n-1)*2 => n= 101
hence, sum = 101/2*[2*100+(101-1)*2] = 20,200