Explanation

INPUT-OUTPUT approach.
Let's choose some nice input values of j and k and see what the answer (output) to the question becomes.
Let's try j = 0 and k = 2
So, the number of even integers that are greater than 0 and less than 2 = 0
So, in this case, when we input j = 0 and k = 2, the output (answer to our question) is 0

Now we'll check the answer choices to see which one yields an output of 0, when we plug in j = 0 and k = 2

A. \(\frac{(k -j -2)}{2}\) we get: (2 - 0 - 2)/2 = 0 KEEP

B. \(\frac{(k -j -1)}{2}\) we get: (2 - 0 - 1)/2 = 1/2 ELIMINATE

C. \(\frac{(k -j )}{2}\) we get: (2 - 0 )/2 = 1 ELIMINATE

D. \(k -j\) we get: 2 - 0 = 2 ELIMINATE

E. \(k -j -1\) we get: 2 - 0 - 1 = 1 ELIMINATE

Answer: A
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Sandy, there will be (j+2n-j)/2 + 1 number of even integers will be there right?,
that will be n+1, how come you have taken n-1 integers, please explain?

even if we take j=2 and k=8,

there will be 2, 4, 6, 8

since j = k+2n

at n = 3 j=2+6=8, here there are 4 numbers are there n+1 = 3+1= 4.

on that case, if we write n+1 interms of j and k. n+1 = (j-k+2)/2, which is not given in answer!,

In the series 2, 4, 6, 8 here j=2 and k=8

n= 3 as \(k = j + 2*3\).

The numbers between 2 and 8 are 4 and 6 or 2 terms.


The reason you get n+1 terms is because you have added the first term (2) and final term (8). That is incorrect!
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There are two formulas that can be used to find the # of values for any arithmetic sequence.

Inclusive:

# values in arithmetic sequence inclusive = [ (last value - first value)/common difference ] + 1

which is the same as

(last value - first value + common difference)/common difference

(k - j + d)/d

(k - j + 2)/2



Exclusive:

# values in arithmetic sequence inclusive = [ (last value - first value)/common difference ] - 1

(last value - first value - common difference)/common difference

(k - j - d)/d

(k - j - 2)/2

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