I am sure there would be an easier method or a formula to solve the problem below? Q: If A equals the sum of the even integers from 2-20, inclusive, and B equals the sum of the odd integers from 1-19, inclusive, what is the value of a-b ? A) 1 b) 10 C) 19 D) 20 E) 21

There is an easy method. Formula might be more time-taking. a = 2+4+6+......20 b = 1+3+5+......19 subtract b from a, it will be (2-1)+(4-3)+(6-5)+.......(20-19) so a-b = 1+1+1+......1 = 10

right .. this way works... but i mean isn't this method too time consuming to be used on gmat?... it is almost same as adding all the numbers and then finding out the difference ?

use Arithmetic Progression formula: Sum = n/2[a +l] where n = number of integers; a= first number; b=last number in the sequence solve for the 2 groups and subtract...the answer is 10.

a(n)=a1+(n-1)d S=(n/2)(a1+an) d=difference between the consecutive numbers for the first series a=2+4+6+...20 d=2 a1=2 an=20 a(n)=a1+(n-1)d 20=2+(n-1)2 =>n=10 S=(10/2)(2+20) =>S(a)=110 same for b =>n=10 and S=(10/2)(1+19) =>S(b)=100 S(a)-S(b)=110-100=10 You should be able to do most of the math in your head. R

If you have consecutive integers (or consec even / consec odd), you can find the sum by finding: 1) the average of the first and last terms, and 2) the # of terms and then multiplying those two together. So, for even from 2-20 1) avg = 22/2 = 11 2) # = (20/2) - (2/2) + 1 = 10-1+1 = 10 (Note: need to divide by 2 b/c only counting the even terms) multiply to get 110 For odd from 1-19 1) avg = 20/2 = 10 2) # = 10 (can tell b/c there were 10 terms for 2 to 20 - same as 1 to 19) multiply to get 100 difference = 10

Wow! what a great thread. I was certain when I saw this question there was only one way to solve it: Using the formula for adding consecutive integers. I liked Rajesh's solution. Cheers guys, I'll remember the two new tricks I learnt today to deal with arithmetic progressions...

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