What is sum of all even integers from 650 to 750, inclusive?

Even before we start off solving for the sum, let us eliminate answer option A. The sum of even integers from 650 to 750 can never be as small as 3500.
If you are hard pressed for time, you could do something similar with answer options D and E as well, since the sum cannot be as big as 70000. How does one be sure of that? Simple, ask yourself, “Do I have 100 values here, each of which are in and around 700?”

The sequence in discussion is the sequence of even integers from 650 to 750, inclusive, which is
650, 652, 654, …….. 746, 748, 750.

This is an equally spaced sequence of values. In an equally spaced set of values,

Mean = \(\frac{(First Element + Last element) }{ 2}\)

Therefore, mean of the given sequence = \(\frac{(650 + 750) }{ 2}\) = \(\frac{1400 }{ 2}\) = 700

By definition, Mean = \(\frac{Sum of values }{ Number of values.}\)
We can modify the above equation to find the sum as, Sum of values = Mean * Number of values.

650 = 2 * 325 and 750 = 2 * 375.
Number of integers from 325 to 375 = (375 – 325) + 1 = 51.

Therefore, sum of all even integers from 650 to 750 = 700 * 51 = 35700.

The correct answer option is C.

Correct solution found, but you mean option C, not option B.

Using the formula of the sum numbers we get:

arithmetic average = (650 + 750)/2 = 700
Quantity of terms = (750 - 650)/2 + 1= 51 (we divide by 2 in order to only get the even terms of the sequence).

(Avg) * (# of Terms) = 700 * 51 = 35,700.

IMO C.