This is not a question. It will probably be flagged before this solution is seen. In future, I will not volunteer a solution until you have asked the question.
You should know by now how to ask a question on these forums.
The implied query seems simple, though:
public Enumerable<int> NumberPattern(int n){
int baseNum = 5;
for(int i = n; i < 2*n; i++){
yield baseNum * i
}
for (int i = 0; i < n; i++){
yield baseNum * (n + i);
}
int[] result = new int[n]{
for (int i = 0; i < n; i++){
result[i] = baseNum * (n + i);
}
return result;
}
</int>
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