Sum of first 50 even numbers is..

[mynameisdong]

Hi all,

Just to make sure I understand the question -

We [(200-102) / 2] + 1 = 50 to get the number of even numbers within 102 - 200 inclusive correct?

If they asked for all the odd numbers, how would you go about finding how many odd numbers there are?

Only way I could think of was, (200-102)+1 to get the total number of numbers, and subtract that by 50 (# of even numbers..).

any other way to get this? Just wanted to know how to do this inside and out.

thanks!

[RonPurewal]

Hi all,

Just to make sure I understand the question -

We [(200-102) / 2] + 1 = 50 to get the number of even numbers within 102 - 200 inclusive correct?

If they asked for all the odd numbers, how would you go about finding how many odd numbers there are?

Only way I could think of was, (200-102)+1 to get the total number of numbers, and subtract that by 50 (# of even numbers..).

any other way to get this? Just wanted to know how to do this inside and out.

thanks!

i'll tell you the way i do it, which is much more intuitive and much less formula-based (i can't remember a formula to save my life): i just transform the numbers until i have gotten them to turn into a list of consecutive integers 1, 2, 3, ..., N.

so, with the evens from 102 to 200:
102, 104, 106, ..., 200

divide by 2 (since they're separated by 2; we want consecutive integers, which are separated by 1)
--> 51, 52, 53, ..., 100

subtract 50 (since we want the first integer to be 1)
--> 1, 2, 3, ..., 50

so there are 50 of them.

if you want to try this technique out, give it a shot on your own question with the odds.

[jatin.harsh]

Hi,

I can make out only sum of 50 even numbers (2,4,6,...,100) can be 2550.

Question stem says sum of first 50 even numbers. Should n't it be first 50 positive even numbers?

Regards
Jatin

[RonPurewal]

Hi,

I can make out only sum of 50 even numbers (2,4,6,...,100) can be 2550.

Question stem says sum of first 50 even numbers. Should n't it be first 50 positive even numbers?

Regards
Jatin

well, remember that the transcriptions you see here are sometimes less than completely faithful to the original.
i went and looked in an archive of screenshots, and the actual gmat prep problem statement says "the first 50 positive even integers". i will edit the original post accordingly.

[sachin.w]

use formula:

sum = n/2[2a + (n-1)d ]

a first term, n no of terms.
d difference..

works like charm :D

Guess remembering AP's sum and last term formula will anytime help. . could solve many gmat prep questions using these formula :D

[jnelson0612]

use formula:

sum = n/2[2a + (n-1)d ]

a first term, n no of terms.
d difference..

works like charm :D

Guess remembering AP's sum and last term formula will anytime help. . could solve many gmat prep questions using these formula :D

Yes, I agree. I see this problem and immediately think of the relevant consecutive integer formula. I guess the takeaway here is that there are a variety of ways to think about these problems. :-)