tapanmittal wrote:For any positive integer n,the sum of the first n positive numbers equals n(n+1)/2.What is the sum of all the even integers between 99 and 301?
Ans-20,200
For any positive integer n, the sum of the first n positive integers equals [n(n+1)]/2. What is the sum of all even integers between 99 and 301?
A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150
Solution:
Although a formula is provided in this problem, we can easily solve it using a different formula:
sum = (average)(quantity)
Let's first determine the average.
In any set of numbers in an arithmetic sequence, we can determine the average using the formula:
(1st number in set + last number in set)/2
Remember, we must average the first even integer in the set and the last even integer in the set, which are 100 and 300, respectively. So we have:
(100 + 300)/2 = 400/2 = 200
Next we have to determine the quantity. Once again, we include the first even integer in the set and the last even integer in the set. Thus, we are actually determining the quantity of even consecutive integers from 100 to 300, inclusive.
Two key points to recognize:
1) Because we are determining only the number of "even integers" in the set, we must divide by 2 after subtracting our quantities.
2) Because we are counting the consecutive even integers from 100 to 300, inclusive, we must "add 1" after doing the subtraction (because we are including both 100 and 300 in the calculation).
quantity = [(300 - 100)/2] + 1
quantity = (200/2) + 1
quantity = 101
Finally, we can determine the sum.
sum = 200 x 101
sum = 20,200
Answer:
B
