For any positive integer n,the sum of first n positive integer equals n(n+1)/2.What is the sum of all the even integers between 99 and 301?
Please solve with explanation
Please solve with explanation
The question asks for the sum of the even integers from 100 to 300, inclusive.ruplun wrote:For any positive integer n, the sum of 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
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.
