Total number of terms (n) = 11 [odd]
∴ Middle most term=(n+1)2th term=(11+12)th term=6th term
Hence, a6=30
⇒a+(6−1)d=30 [∵ an=a+(n−1)d]
⇒a+5d=30...(i)
Sum of terms of an AP,
Sn=n2[2a+(n−1)d]
∴ S11=112[2a+(n−1)d]
=112(2a+10d)=11(a+5d)
=11×30=330 [from eq(i)]