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

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

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New post 02 Apr 2018, 01:07
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C
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Re: What is the sum of all the even integers between 99 and 401 ?  [#permalink]

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New post 02 Apr 2018, 01:17
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


As there are rules for straightforward calculation of the sum of arithmetic sequences, we'll use them.
This is a Precise approach.

The first number in the sequence is 100, the last is 400.
As there are a total of (401-99)/2 = 302/2=151 numbers, our sum is
(100+400)*151/2 = 250*151 = about the middle between 250*200 = 50,000 and 250*100 =25,000
That is, about 37,500.

(C) is our answer.
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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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New post 02 Apr 2018, 11:21
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1
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]

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New post 08 Apr 2018, 03:30
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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New post 09 Apr 2018, 09:02
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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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New post 30 Apr 2018, 08:42
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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New post 03 May 2018, 15:34
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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:
\(n =\) \((\frac{LastTerm-FirstTerm}{2}+1)\)
\((\frac{400-100}{2}+1)=150 + 1 =\) \(151\)


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


\(=37,750\)

Answer C

Hope that helps. :-)
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Re: What is the sum of all the even integers between 99 and 401 ?  [#permalink]

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New post 04 May 2018, 03:02
1
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:
\(n =\) \((\frac{LastTerm-FirstTerm}{2}+1)\)
\((\frac{400-100}{2}+1)=150 + 1 =\) \(151\)


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


\(=37,750\)

Answer C

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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New post 07 May 2018, 09:23
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

Answer: C
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Re: What is the sum of all the even integers between 99 and 401 ? &nbs [#permalink] 07 May 2018, 09:23
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What is the sum of all the even integers between 99 and 401 ?

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