Let 1 to 20 are placed in any around a circle. Then the sum of some 3 consecutive numbers must be at least
(a) 30
(b) 31
(c) 32
(d) none
My approach
each number participates in exactly three triplets. Since the numbers add up to 210, the triplets add up to 630 There are 20 of them, so the average triplet has sum 31.5. but it has to be an integer so the ans should be 32
But my doubt is {1,2,3} be also a triplet.At that time it will violate my solution.
Your argument is correct, if a little poorly worded. Each triplet must have a sum that is an integer, and if a set of integers have an average of $31.5$ some of them have to be at least $32$. As N. F. Taussig says, we are looking for the maximum sum of three consecutive numbers, then taking the minimum of that over all arrangements. However if $32$ is correct, so are $30$ and $31$ because whatever number is at least $32$ is also at least $31$.