The sum 1 + 3 + 9 + 27 + … of a geometric progression first exceeds 1000 when the number of terms is:

With $a=1$ and $r=3$ the sum of n terms is (3ⁿ − 1)/2, so we need 3ⁿ > 2001. Since $3^6=729$ and $3^7=2187$, the first n that works is 7. (6 gives the sum 364, still short; 5 gives 121; 8 works too but is not the first.) Listing them: 1, 4, 13, 40, 121, 364, 1093 — the 7th total is the first above 1000.