Which of the following is the sum of all the even numbers between 1 an

Question

Which of the following is the sum of all the even numbers between 1 and 99, inclusive?

A. 2550

B. 2450

C. 2600

D. 2499

E. 2652

EMPOWERgmatRichC · Accepted Answer

Hi hiteshahire22,

This question can be solved with "bunching."

We're asked for the SUM of all of the EVEN integers from 1 to 99, so we really want the sum of 2, 4, 6, 8.....98.

If you take the smallest and biggest numbers, you have...
2+98 = 100

If you take the next smallest and next biggest numbers, you have...
4+96 = 100

This pattern will continue on, so we're going to have a bunch of 100s, BUT we have to be careful to make sure that we don't miss a number if it's "in the middle" and doesn't get paired up. Since we know that the sum of each pair is 100, we can 'jump ahead' to find the last few pairs...

44+56 = 100
46+54 = 100
48+52 = 100
There IS a middle number: 50; this number does NOT get paired up.

Since 48 is the 24th even integer, we know there are twenty-four 100s + one 50. 2400+50 = 2450

Final Answer:  B

GMAT assassins aren't born, they're made,
Rich

ENGRTOMBA2018 · Answer

Sum of an arithmetic series = Sn =  \frac{n(2a+(n-1)d)}{2} , where n = number of terms = number of even numbers between 1 and 99 = 49, d = difference between 2 consecutive even terms = 2, a = 1st term of the series = 2.

Thus Sn =  \frac{49(2*2+(49-1)2)}{2}  = 49*50 = a bit less than 50*50 = a bit less than 50*50 but with 0 as the unit's digit =B is the only option satisfying this.

B , 2450 is the correct answer.

Turkish · Answer

We need to find two things- Number of even terms and then Sum of all the terms.

Number of even terms= last even number - 1st even number/ spacing between two even numbers + 1. so, 98-2/2+1 = 49. now Sum of all the terms = Number of even terms * last number + 1st / 2 which is 49*99+1/2 = 2450

AbhimanyuDhar · Answer

There are 2 Formulas we need to know

1 Sum of even numbers <=n where n is a natural number

in case N is odd then sum is given by {(n-1)(n+1)/4}

in case N is even then sum is given by n(n+2)/4

So in our case since n=99 is odd we have (98*100/4)= 2450

VyshakhR1995 · Answer

49*2=98 
---->49 Numbers
n/2(a1+an)

49(2+98)/2
4900/2
2450

Kinshook · Answer

Asked: Which of the following is the sum of all the even numbers between 1 and 99, inclusive?

Sum = 2 + 4 + ... 98 = 49/2(4 + 48*2) = 49/2 (100) = 49*50 = 2450

IMO B

prateekchugh · Answer

Which of the following is the sum of all the even numbers between 1 and 99, inclusive?

2+4+6..........98

In 1 to 100; total number of odd=50 and total number of even = 50
In 1 to 99; 100, which is even, is excluded; total number of odd=50 and total number of even = 49

Many Ways you can solve this problem:

1) Sum of numbers, S =  \frac{(2*a_1+ (n-1)d)*n}{2}

a_1 = 2, n=49, d=2

Enter the values into the formula

S= 50*49=2450

2) Sum of even number = Average of even number * total even number= ( \frac{1st term+ Last term}{2} ) * n

1st term = 2; Last term = 98; n=49

Insert the values into the equation

Sum of even numbers= ( \frac{2+ 98}{2} ) * 49= 50*49=2450

3) Sum of even numbers= n(n+1); and sum of odd number =  n^2

n=49

Sum of even number = 49 *(49+1)=49*50=2450

Answer is B

A. 2550

B. 2450

C. 2600

D. 2499

E. 2652

MathRevolution · Answer

The range is from 1 to 99.

First even number (a) = 2 ; Common difference (d) = 2 ; Last even number = 98

=>  T_n  = a + (n-1)d

=> 98 = 2 + (n-1)2

=> n = 49

=> Sum =  \frac{n}{2}  *

=> Sum =  \frac{49 }{ 2}  *

=> Sum = 2450

Answer B

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Which of the following is the sum of all the even numbers between 1 an

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