The sum of N consecutive positive integers, the first of which is a, is (a + (a+N-1))N/2.

If this equals 3^11, then either N or N/2 is a factor of 3^11.

Thus, N can only be 3^k where k is between 0 and 11, or 2*3^k where k is between 0 or 10.

But some of these are obviously too large, since N^2 < 2*3^11.

From this equation, N < sqrt(6) * 3^5

Of our candidates, the largest possible N is 2*3^5.

Will this work?

Substituting and solving for a,

(2a+ 2*3^5 - 1)*3^5 = 3^11

a = (3^6 + 1 - 2*3^5)/2 = (3^5 + 1)/2

This is a whole number, so the maximum N is indeed 2*3^5.

The last term is (5*3^5 - 1)/2.

The average term is 3^6/2, so the total sum is indeed 3^11