Question: If the average of 18 consecutive odd integers is 534, what is the least number in the set?
Ok so I know how to do this algebraically, you start with your odd variable which we will call X and then the next one will be X + 2, X + 4, etc until you have 18 terms. Then it will be 18X + 306/18 = 534 and then we can solve, but this method seems rather long. Is this the best way to solve this problem?
There are three sum formulas that you should commit to memory:
1. Sum of the first n consecutive integers: n(n+1)/2
2. Sum of the first n consecutive even integers: n(n+1)
3. sum of the first n consecutive odd integers: n^2
So, back to the problem, we have: sum / number of odd integers = [n + (n+2) + (n+4) + ... + (n+34)] / 18 = 534. We can reduce this to 18n + (2 + 4 + 6 + ... + 34) / 18 = 9612 ==> 18n + (2+ 4 + ... + 34) = 173016. We can simplify the sum in the parenthesis using rule 2 from above: 17 x 18 = 306. So we have 18n + 306 / 18 = 534. And now you can solve in the usual way.
Users browsing this forum: Yahoo [Bot] and 3 guests
You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot post attachments in this forum
GMAT(TM) and GMAT CAT (TM) are registered trademarks of the Graduate Management Admission Council(TM). The Graduate Management Admission Council(TM) does not endorse, nor is affiliated in any way with the owner or any content of this site.