The arithmetic mean of the even integers from 200 to 300

Lets consider a smaller range:
(1) 200 to 204
even numbers: 200, 202, 204; Avg = 202
Odd numbers: 201, 203; Avg = (201+203)/2 = 202
Difference = 0

(1) 200 to 210
even numbers: 200, 202, 204, 206, 208, 210; Avg = (204+206)/2 = 205
Odd numbers: 201, 203, 205, 207, 209 Avg = (201+203+205+207+209)/5 = 205
Difference = 0

Ans: C.

Evenly spaced sequence

The arithmetic mean of an arithmetic sequence is
\(\frac{FirstTerm + LastTerm}{2}\)

Even integers, arithmetic mean:
\(\frac{200+300}{2} = 250\)

Odd integers, arithmetic mean:
\(\frac{201 + 299}{2} = 250\)

Difference: 0

Answer C

Small sample
As posters above demonstrate, to use a small sample is often shrewd.
Yet another way to take a small sample:

Replicate the pattern of the numbers in the prompt
• begin and end with even numbers
• if you're not sure, use integers that end in the same units digit (here, 0), and
• if you're going to do all the math, choose small numbers

The arithmetic mean of the even integers from 0 to 10, inclusive
\(0, 2, 4, 6, 8, 10\), where \(A =\frac{S}{n} = \frac{30}{6} = 5\)

Arithmetic mean of odd integers from 0 to 10, inclusive
\(1, 3, 5, 7, 9\), where \(A=\frac{S}{n} = \frac{25}{5} = 5\)
Difference: 0

Answer C

Let's cut out wasted time on arithmetic here. This can simply be done with thought.

The amount of even integers will be two more integers than the odd range. Additionally, the even range will have an amount of 2 more than the odd range (201-200 & 300-299)

Therefore,

avg = sum / amount

no change = X+2 / X+2

Even integers from 200 to 300 inclusive : 200 , 202, 204, .... 300 = 50 terms.

Artihemtic Mean : \(\frac{200 + 300 }{ 2}\) = 250 ( As they form an A.P.)

Odd integers from 200 to 300 inclusive : 201 , 203, 205, .... 299 = 50 terms.

Artihemtic Mean : \(\frac{201 + 299 }{ 2}\) = 250 ( As they form an A.P.)

Difference : 250 - 250 = 0

Answer C

MathRevolution for number of terms formula is last term - first term / difference . When is +1 added or not added?
Not sure about this. Can you please help.

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