shasadou wrote:

The sum of the first 100 even positive integers – divided by the sum of the next 100 even positive integers – is equal to which of the following?

A. 1/3

B. 101/ 301

C. 51/151

D. 1/2

E. 1

Ans(B) 101/301

Given that, p = (2+4+6.....+200 ) [First 100 Terms] / (202+204....+400) [Next 100 Terms]

p= 2(1+2+...+100) / 2 ( 101+ 102+200)

P= (1+2+....+100) / (101+102+...+200) ( Cancelling out 2)

After this we can, we can solve this using 2 methods.

Method1:To calculate the sum of a continuous series, with a same difference between any two continuous numbers, simply take the average of first & last number and multiple it by the number of terms. For ex : 4+7+10+13= (average of 13&4, multiplied by 4) = 8.5*4 = 34

Now in the original problem,

p = Average of (1&100)*100 / Average of (101&200)*100

p=(1+100)/(101+200) (Cancelling out 2, not shown here)

p=101/301 Answer.

Method2: [This is a formula dependent approach and lengthy one also]

For Equation, P= (1+2+....+100) / (101+102+...+200)

We can use the formula for the sum of first "n" natural numbers : S= n(n+1)/2. But this formula works when the series starts from 1.

So, using this formula, we can calculate the sum for the numerator part of the above equation.

But for denominator, first we need to calculate the some of 200 numbers and then subtract the sum of first 100.

And this double calculation makes this approach slightly lengthy.

Thanks!