Set C is a set of positive integers where 4 is least. If the average

Hi septwibowo
The question is asking How many integers i.e. count of no of elements in set C. Hence it is not about whether the elements are integer or not. Even if they are not, then also we will get a count.

First term is \(4\), let the last term be \(T_n\) and let the number of element be \(n\). Given \(T_n-4=Average=\frac{Sum}{n}, or Sum = n(T_n-4)\)

Statement 1: Approach 1:

this implies that the set is an AP with common difference of \(1\). Sum of an AP series \(= \frac{n}{2}(first term+Last term)\)

or \(n(T_n-4)=\frac{n}{2}(T_n+4)\) or \(T_n=12\)

So the element will be from \(4\) to \(12\) i.e \(9\) Sufficient

Approach 2: as the elements are consecutive, the average will be the middle number and as average is an integer, hence number of elements will be ODD. so you can simply list down numbers starting from 4 to arrive at a set 4,5,6,7,8,9,10,11,12, with 8 as average = range (12-4)

Statement 2: implies \(T_n-4=8\) or \(T_n=12\), so Set C can have only two element {4,12} or can be {4,5,6,7,8,...12}. Hence Insufficient

Option A