Bunuel wrote:
If x is the sum of the even integers from 200 to 600 inclusive, and y is the number of even integers from 200 to 600 inclusive, what is the value of x + y?
(A) 200*400
(B) 201*400
(C) 200*402
(D) 201*401
(E) 400*401
Kudos for a correct solution.
VERITAS PREP OFFICIAL SOLUTION:There are various ways of getting the answer here. We will use the concepts we learned last week.
The given sequence is 200, 202, 204, … 600
It is an arithmetic progression. What is the total number of terms here?
You can use one of two methods to get the number of terms here:
Method 1: Using Logic
In every 100 consecutive integers, there are 50 odd integers and 50 even integers. So we will get 50 even integers from each of 200 – 299, 300 – 399, 400 – 499 and 500 – 599 i.e. a total of 50*4 = 200 even integers. Also, since the sequence includes 600, number of even integers = 200 + 1 = 201
Method 2:Recall that in our
arithmetic progressions post, we saw that the last term of a sequence which has n terms will be first term + (n – 1)* common difference.
\(600 = 200 + (n – 1)*2\)
\(n = 201\)
Hence \(y = 201\) (because y is the number of even integers from 200 to 600)
Let’s go on now. What is the average of the sequence? Since it is an arithmetic progression with odd number of integers, the average must be the middle number i.e. 400.
Notice that since this arithmetic progressions looks like this:
(n – m), … (n – 6), (n – 4), ( n – 2), n, (n + 2), (n + 4), (n + 6), … (n + m)
We can find the middle number i.e. the average by just averaging the first and the last terms.
\(\frac{(n – m) + (n + m)}{2} = \frac{2n}{2} = n\)
\(Average = \frac{(200 + 600)}{2} = 400\)
Sum of all terms in the sequence = x = Arithmetic Mean * Number of terms = 400*201
\(x + y = 400*201 + 201 = 401*201\)
Answer (D)
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