The sum of the first 50 positive even integers is 2550. What is the

# of even terms (multiples of 2) in the range from 102 to 200, inclusive is \(\frac{(last - first)}{2} + 1=\frac{(200-102)}{2}+1=50\);

\(Mean = \frac{(last+first)}{2}\);

\(The \ sum = (mean)*(number \ of \ terms) = \frac{(200+102)}{2}*50 = 7,550\).

Answer: B.
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I had derived a formula for calculating the summation of even/odd numbers in a series, when I was in high school.

Sum of EVEN numbers in a series from 1 -> n (where n is the LAST EVEN number in a series) = n*(n+2)/4
Sum of ODD numbers in a series from 1 -> n (where n is the LAST ODD number in a series) = (n+1)*(n+1)/4

This works perfect for any series starting from 1. Using the above formula we can solve the problem:

200*202/4 - 100*102/4 = 7550

Please give Kudos if this was helpful.

The sum of the first 50 positive even integers is 2550.
In other words, 2 + 4 + 6 + 8 + ...+ 98 + 100 = 2550

We want the sum: 102 + 104 + 106 + . . . 198 + 200
IMPORTANT: Notice that each term in this sum is 100 GREATER than each term in the first sum.
In other words, 102 + 104 + 106 + . . . 198 + 200 is the SAME AS...
(100 + 2) + (100 + 4) + (100 + 6) + ... + (100 + 98) + (100 + 100)
We can rearrange these terms to get: (100 + 100 + ... + 100 + 100) + (2 + 4 + 6 + 8 + ...+ 98 + 100)

IMPORTANT: There are 50 100's in the red sum, and we're told that the blue sum = 2550

So, our sum = 50(100) + 2550
= 5000 + 2550
= 7550

Answer:

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My solution is:

First 50 even integers:
2
4
6
8
<...>

Integers from 102 to 200
102
104
106
108
<...>

We notice that each integer from the second set is 100 more than the respective integer in the first set. Since we have 50 even integers from 102 to 200, then:
2550+(100*50)=7550.

Are there any other ways to solve the prob? Thanks!

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Please kudos me if you find my post useful

Sum of even integers from 102-200(both included)

No of terms = (200-102)/2 + 1 = 50 (+1 because both are included)

If you observe, this is AP (Arithmetic progression): 102,104,106,......,200
Sum of n terms in AP Formula, Sum = n/2[2a + (n-1)d]
where,
n = no of terms = 50 in this case.
a = starting number = 102
d = difference between AP = 2

So, Sum = 50/2[2*102 + (50-1)2]

Solving this, we can get the answer.

This is one more solution for this kind of problems.

Sum of even integers from 102-200(both included)

No of terms = (200-102)/2 + 1 = 50 (+1 because both are included)

If you observe, this is AP (Arithmetic progression): 102,104,106,......,200
Sum of n terms in AP Formula, Sum = n/2[2a + (n-1)d]
where,
n = no of terms = 50 in this case.
a = starting number = 102
d = difference between AP = 2

So, Sum = 50/2[2*102 + (50-1)2]

Solving this, we can get the answer.

This is one more solution for this kind of problems.

Saurabhricha,
what would be the formula for odd nbrs?

Sum = Avg * number of items

Avg = (200 + 102) / 2 = 302 / 2 = 151

No. of items = 200-102 = 98/2 (Only the even numbers so divide by 2) = 49 + 1 (add 1 the last item) = 50

Sum = 151 * 50 = 7550

ANS : B

Even easier method:

First 50 even integers are 2, 4, 6, ... , 100. Sum is stated as 2550.
Sum of even integers between 102 and 200 are 100 more than each integer in the previous sum (102 is 100 more than 2, etc.).

So...when this happens 50 times, it's 50*100 = 5000, 5000 + 2550 = 7550.

No need for formula, just simple number pattern here.

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My GMAT Experience: 670-to-760-q49-v44-ir-8-awa-6-0-no-stone-unturned-for-8-months-200051.html

I am totally confused. in the math book, the formula for the first n even number is n(n+1)

and here there is another formula used, mean*# of terms


Bunuel please explain!!!

The sum of n first positive even numbers is n(n+1) --> the sum of first 50 positive even numbers is 50*51 = 2,550
The sum of evenly spaced set is (mean)*(# of terms) --> the sum of first 50 positive even numbers is (2+100)/2*50 = 51*50 = 2,550.
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2 + 4 + 6 + 8 ................... + 100 = 2550 ................. (1)

102 + 104 + 106 + 108 ............ + 200 = ?? ................... (2)

Just note that we have to add 100 to each term of series (1) to get the corresponding terms in series (2)

(2+100) + (4+100) + (6+100) + (8+100) ................ + (100+100)

This 100 ha to be added 50 times, so result of series (2) would be

2550 + 50*100 = 7550

Answer = B

For any EVENLY SPACED SET:
Count = \(\frac{(biggest - smallest)}{(increment)}\) + 1
Average = median = \(\frac{(biggest + smallest)}{2}\)
Sum = (count)(average)
The INCREMENT is the difference between successive values.


Even numbers between 102 and 200, inclusive:
Here, the integers are EVEN, so the increment = 2.
Count = \(\frac{(200-102)}{2}\) + 1 = 50.
Average = \(\frac{(200+102)}{2}\) = 151.
Sum = (50)(151) = 7750.




Alternate approach:

102, 104, 106......196, 198, 200.

The median is halfway between 102 and 200:
\(\frac{(102+200)}{2}\)= 151.
Of the 100 integers between 101 and 200, inclusive, half are odd, half are even.
Thus, in the set above, there are 25 even integers to the LEFT of 151 and 25 even integers to the RIGHT of 151, for a total of 50 even integers.

Add successive PAIRS of integers, working from the OUTSIDE IN:
102+200 = 302.
104+198 = 302.
106+196 = 302.

Notice that the sum of each pair = 302.
Since there will be 25 of these pairs, the following sum is yielded:
25*302 = 7550.


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SUM OF N TERMS = \(\frac{n}{2} (2a+(n-1)d)\)

\(a\) = 102 the first term
\(d\) = 2 distance
\(n\) = 50 how many terms to add up


\(\frac{50}{2} (2 (102)+(50-1)2)\) = 7550



another approach Sum = average x number of terms

Average = \(\frac{(200 + 102)}{2} = 151\)

Number of terms \(\frac{(200-102)}{2} + 1 = 50\)

Sum of Even integers 151 *50 = 7,550

[[(200-102)/2 ] +1]*(200+102)/2

The GMAT does not expect the testtaker to remember formulae for the sum of even or odd integers.

The hint is provided in the question. The question has already provided the sum of the first 50 even integers. We are expected to use this information in our solution.

The sum of the first 50 even integers is (2 + 4 + 6 + ............. + 98) = 2550 (given).

We are asked to determine the sum of the next set of even integers (102 + 104 + 106 + .... + 198).

Note that this second set simply includes elements that have an added amount of 100 for each of the elements in the previous set. So we have an additional amount of 100 multiplied by 50. The quantity 50 arises from noting that there are 50 even integers in the second set, as there were in the first set.

The answer is simply: 2550 + 50*100 = 7550.

ANSWER: B

Note: There is no need to remember or use formulae for sums of integers, whether odd or even. All information has already been provided in the question.

There are 50 even integers between 2 and 100 (inclusive) and there are another 50 even integers between 102 and 200 (inclusive).

It is given that the sum of the 1st 50 positive even integers is 2550.
The sum of the first 100 even integers= 100x101= 10100. (The sum of the first m even whole numbers = m(m+1) and the sum of the first m odd numbers= m^2)

Hence the sum of even integers from 102 to 200= 10100-2550= 7550. Correct answer is B.

If you aren't sure how to obtain the number of terms between 102 and 200 or how to get the 50th term in the first interval, use the following formula applicable to arithmetic progressions: An=A1+ (n-1)d; Where An is the nth term, A1 is the 1st term, n is the number of terms and d is the difference between every two successive terms (and that is always the same).

For example to find the 50th term in the 1st interval: A50=2+ (50-1)x2=100.

Also to find how many terms exist between 102 and 200: An=102+ (n-1)x2=200; solving for n, you get n=50.