Answer:
You need at least 37 terms to have a sum of 2000 or more
Step-by-step explanation:
,
S(n):=3+6+9+12+⋯+3n=3⋅n(n+1)2
This is approximately 32⋅n2.
It should be around 2000, that is n2 should be around 2/3⋅2000≈1334. Using calculator, its square root is ≈36.52 and if you sustitue n=36 to S(n) gives S(36)=1998.
So since 36 is not enough you need 37 terms to have a sum of 2000 or more
