Step 1: List the terms with their position number $n$: $T_1=3, T_2=6, T_3=11, T_4=18, T_5=27$.
Step 2: Since the first differences (3, 5, 7, 9) themselves form an arithmetic progression, the sequence is a quadratic in $n$: assume $T_n = an^2+bn+c$.
Step 3: Substitute three known terms to form equations: \[a+b+c=3,\quad 4a+2b+c=6,\quad 9a+3b+c=11\] Subtracting consecutive equations gives $3a+b=3$ and $5a+b=5$. Subtracting these: $2a=2 \Rightarrow a=1$, then $b=0$ and $c=2$.
Step 4: So the closed-form rule is $T_n = n^2+2$. Checking: $T_1=1+2=3$, $T_2=4+2=6$, $T_3=9+2=11$, $T_4=16+2=18$, $T_5=25+2=27$ - all match.
Step 5: For the next term, $n=6$: \[T_6 = 6^2+2 = 36+2\]
\[\boxed{T_6 = 38}\]