We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

The set \(\{2, 4, 6, \dots, n\}\) contains the positive consecutive even integers from 2 through \(n\). When one of the integers from the set is removed, the average of the remaining integers in the set is 28. What is the least possible value of \(n\)?

I got that the sum of the set is \(n+n^2\) but I don't know if that will help me.

DanielCai Jul 3, 2018

#1**0 **

I believe the smallest value of n = 52 as follows:

The set {2, 4, 6, 8,.........52}. The 52 is the 26th and last term of the set. If you add up all the terms of the set: 2 + 4 + 6 +.............+52 =702. If you remove the first term, or 2, you will have left 25 terms that total 700. The average of the remaining 25 terms of the set is (702 -2) / 25 = 28.

Guest Jul 4, 2018