farooq wrote:For any positive integer n, the sum of the first n( n+ 1)/n positive integers equals 2 . What is the sum
of all the even integers between 99 and 301 ?

(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150

B

Please explain.

quite a few way: the numbers are: 100 102 104 .... 300 since the numbers are equally spaced you may find the mean and multiply it by total numbers.

If you know the AM formula : Tn = a + (n-1)*d here 300 = 100 + (n-1)*2 or n = (300-100)/2 + 1 = 101
Sum : n/2{first term + last term} = (101/2)*400 = 20200

Using the above formula : need to convert these number from 1..n
S = 100 + 102 + 104 + ... + 300
= 100 + 100+2 + 100+4 + ...+ 100+200
= 100*101 + (2+4+6+..+200)
= 10100 + 2*(1+2+3+..100)
= 10100 + 2*100*101/2
= 20200

total no. of terms between 100 and 300 = 200/2 + 1 = 101

Mean = (100+300)/2 = 200

Sum of even nos. = 101*200 = 20,200

Would someone please provide a concise breakdown of this problem. I am having a hard time understanding. Based on the question stem I don't get any of the answers, nor do I follow the explanations listed here or especially in the back of the OG.

My process.... N= 100 ----- (300-100)/2

100(100+1) = 10100/2 = 5050

jk2010 wrote:Would someone please provide a concise breakdown of this problem. I am having a hard time understanding. Based on the question stem I don't get any of the answers, nor do I follow the explanations listed here or especially in the back of the OG.

My process.... N= 100 ----- (300-100)/2

100(100+1) = 10100/2 = 5050

Sum of evenly spaced integers is given by - no of terms * Mean of the terms

In the problem above, first even integer is 100 and last even integer is 300

no of terms = 300-100)/2 + 1 = 101.

Mean = (last term + first term)/difference between consecutive term

= 300+100/2 = 200

hence Sum = 101 * 200 = 20200

farooq wrote:For any positive integer n, the sum of the first n positive integer equals n( n+ 1)/2. What is the sum
of all the even integers between 99 and 301 ?

(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150

B

Please explain.

We are given that 1 + 2 + 3 + ... + n-1 + n = n(n+1)/2 and asked to find 100 + 102 + ... + 298 + 300


Compare the two expressions and you will notice that we can write the second as 2(50 + 51 + ... + 149 + 150 ). Why do we do this? Simply to make it look more like the first expression so that we can use the formula provided.

Of course, we would like to start the sum with 1, so we could write

2(50 + 51 + ... + 149 + 150) = 2(1 + 2 + 3 + ... + 150) - 2( 1 + 2 + ... 49)

=2 (150)(151)/2 - 2(50)(51)/2 = 50((3(151) -51)= 50(453 - 51) = 50(402) = 20100


Do read the alternative solutions, as they are faster if you know the technique

jk2010 wrote:Would someone please provide a concise breakdown of this problem. I am having a hard time understanding. Based on the question stem I don't get any of the answers, nor do I follow the explanations listed here or especially in the back of the OG.

My process.... N= 100 ----- (300-100)/2

100(100+1) = 10100/2 = 5050

The question asks for the sum of the even integers from 100 to 300, inclusive.

Ignore the formula given. Instead, use the following:

Sum = (number of integers) * (average of biggest and smallest)

To count the number of evenly spaced integers in a set:

Number of integers = (Biggest - Smallest)/(distance between each successive pair) + 1

Since we're adding only the even integers, the distance between each successive pair is 2.
Thus, the number of even integers from 100 to 300 = (300-100)/2 + 1 = 101.
Average of biggest and smallest = (300+100)/2 = 200.
Sum = (number of integers) * (average of biggest and smallest) = 101*200 = 20,200.

The correct answer is B.

Thanks Mitch and others who took the time to answer.

The formula given was what caused my confusion. The question makes much more sense now!