If we know the sum of first n terms and the sum of first (n−1) terms, then:
an=Sn−Sn−1
This is because when we subtract the sum of first 10 terms from the sum of first 11 terms, we get just the 11th term!
Using our formula with n=11:
S11=3×(211+1−2)
=3×(212−2)
=3×(4096−2)
=3×4094
=12282
Using our formula with n=10:
S10=3×(210+1−2)
=3×(211−2)
=3×(2048−2)
=3×2046
=6138
Now we can find a11:
a11=S11−S10
=3×(212−2)−3×(211−2)
=3×(212−2−211+2)
=3×(4096−2−2048+2)
=3×(2048)
=6144
Answer: a11=6144