First, you need to memorize the formula for the sum of arithmetic progression. Second, you need to know how to count number of terms. Third, practice and practice.

please tell the shortcut method to solve this question.

S(300) = 2+..........+298

for number of terms, l = a+(n-1)*2
298 = a+(n-1)*2 = 2+(n-1)*2 => 296/2 = n-1 => n=149

for sum,

S(300) = n/2 * (a+l) = (149/2)*(298+2) = 149*150 = 22350


Answer is B!

there are three ways to do it ....

(I) If you know that Sum of first n integers is \(\frac{n(n+1)}{2}\)
Sum = \(2+4+6+...+300) = 2(1+2+3....+150)= 2 *\frac{150*151}{2}=150*151=22650\)

(II) If you know that Sum of first n integers is \(\frac{n(n+1)[}{fraction]\)
Now we have \([fraction]300/2}=150\) terms till 300, inclusive.
Sum = \(2+4+6+...+300 = 150*151=150*151=22650\)

(III) since it is an AP. the sum will be equal to Number of integers* average
so \(150 * \frac{(300+2)}{2} = 150*151 = 22650\)

Now subtract 300 from each result as the answered has to be exclusive of 300, that is 300 is not be included in total..
Therefore answer is 22650-300=22350

B

To know more about Arithmetic progressions
https://gre.myprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048

According to the initial problem, the answer is B. Also it seems you are doing inclusive of n=300, while the prompt states NON-inclusive.

Yes, thank you.
I had included 300 in each case.
B will be the correct answer.

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