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What is the sum of all the even integers between 99 and 401 ?

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Re: What is the sum of all the even integers between 99 and 401 ? [#permalink]
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Bunuel wrote:
What is the sum of all the even integers between 99 and 401 ?

(A) 10,100
(B) 20,200
(C) 37,750
(D) 40,200
(E) 45,150

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Re: What is the sum of all the even integers between 99 and 401 ? [#permalink]
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The first term of the sequence is=100
The last term of the sequence is=400

Therefore, Mean=(100=400)/2= 250
Sum= Mean*number of terms = 250*151 = 37,750.

Hence, C.
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Re: What is the sum of all the even integers between 99 and 401 ? [#permalink]
urvashis09 : Can you give me some insight as to why the mean is calculated and how the number of terms 151 is arrived at?
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Re: What is the sum of all the even integers between 99 and 401 ? [#permalink]
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narendran1990 wrote:
urvashis09 : Can you give me some insight as to why the mean is calculated and how the number of terms 151 is arrived at?

The sum of any given set of "consecutive integers" (or any evenly spaced data set) can also be calculated by multiplying mean with the number of integers in that set. You can also derive this formula: Sum= [(X1 +Xn)/2]*n where n = number of consecutive integers in the set.

Hence, in the given set, the first even number = 100, last even number = 400.
Therefore, total number of even numbers = (400-100)/2 + 1 = (300/2)+1 = 150+1 = 151.

And, mean = (400+100)/2 = 500/2 = 250.

Therefore, sum = 250*151 = 37,750.
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Re: What is the sum of all the even integers between 99 and 401 ? [#permalink]
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can anyone explain why cant we use the formula (n/2)(2a+(n-1)d)???
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Re: What is the sum of all the even integers between 99 and 401 ? [#permalink]
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generis wrote:
asfandabid wrote:
can anyone explain why cant we use the formula (n/2)(2a+(n-1)d)???

asfandabid , you can use that formula.
Who said you could not?

In fact, if you look at Bunuel 's post above, in the text with yellow highlight,
he links to excellent posts written by VeritasPrepKarishma .
In one of those posts, here,
she both discusses the formula you like and links to this question as an example.

The posts above do not use the formula that you reference.
Perhaps that is the issue. Either way, below I have answered the question using your formula.
(Note: there are different ways to find number of terms.)

What is the sum of all the even integers between 99 and 401?

$$(\frac{n}{2})(2a+(n-1)d)$$

First Term = $$a = 100$$
Last Term = $$400$$

Number of terms, $$n$$ for even numbers, increment = 2:
$$n =$$ $$(\frac{LastTerm-FirstTerm}{2}+1)$$
$$(\frac{400-100}{2}+1)=150 + 1 =$$ $$151$$

$$(\frac{n}{2})(2a+(n-1)d)=$$
$$(\frac{151}{2})(2(100)+(150)2)$$=
$$(\frac{151}{2}*(200+300))=(\frac{151}{2}*500)=(151)(250)$$

$$=37,750$$

Hope that helps.

Got it. Thanks!!
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Re: What is the sum of all the even integers between 99 and 401 ? [#permalink]
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Bunuel wrote:
What is the sum of all the even integers between 99 and 401 ?

(A) 10,100
(B) 20,200
(C) 37,750
(D) 40,200
(E) 45,150

The number of even integers between 99 and 401 is the same as the number of even integers from 100 to 400 inclusive.

Since sum = average x quantity, let’s determine the average and the quantity.

Average = (100 + 400)/2 = 250

Quantity = (400 - 100)/2 + 1 = 151

Sum = 250 x 151 = 37,750

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What is the sum of all the even integers between 99 and 401 ? [#permalink]
Bunuel wrote:
What is the sum of all the even integers between 99 and 401 ?

(A) 10,100
(B) 20,200
(C) 37,750
(D) 40,200
(E) 45,150

TL;DR

Sum of n terms = n*((a1+an)/2)

# of terms = (400 - 100)/2 + 1 = 200 - 50 + 1 = 151
Avg = (400+100)/2 = 250
Sum = 250 * 151 = 37750

Veritas Prep Official Solution

We can solve this question in multiple ways.

Required Sum = 100 + 102 + 104 + 106 + … + 400

Method 1:

We know the sum of consecutive even integers but only when they start from 2. So what do we do? We find the sum of even integers starting from 2 till 400 and subtract the sum of even integers starting from 2 till 98 from it! Note that we subtract even numbers till 98 because 100 is a part of our series.

How many even integers are there from 2 to 400? I hope you agree that we will have 200 even integers in this range (both inclusive)

Sum of these 200 integers = 200(201)

How many even integers are there from 2 to 98? Now we have 49 even integers here.

Sum of these 49 even integers = 49(50)

What is the sum of integers from 100 to 400? It will be 200(201) – 49(50) = 40200 – 2450 = 37750

Method 2:

100 + 102 + 104 + 106 + … + 400 = 2( 50 + 51 + 52 + 53 + … + 200) (We take out 2 common and find the sum in brackets)

Sum in brackets = 50 + 51 + 52 + 53 + … + 200

We know the sum of consecutive integers but only when they start from 1. So we find the sum of first 200 numbers and subtract the sum of first 49 numbers from it. That will give us the sum of numbers from 50 to 200. Note that we subtract 49 numbers because 50 is a part of our series.

Sum of 1 + 2 + 3 + … + 200 = 200*201/2 = 20100
Sum of 1 + 2 + 3 + … + 49 = 49*50/2 = 1225 (I am doing these calculations here only for clarity. Normally, I would like to carry all these till the last step, then take common, divide by whatever I can etc so that I have very few actual calculations left.)

Therefore, 50 + 51 + 52 + 53 + … + 200 = 20100 – 1225 = 18875

Then 100 + 102 + 104 + 106 + … + 400 = 2*18875 = 37750

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Re: What is the sum of all the even integers between 99 and 401 ? [#permalink]
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Re: What is the sum of all the even integers between 99 and 401 ? [#permalink]
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