n=1∑20anan+11=94 [let common difference of APisd=an+1−an]
=d1n=1∑20(an1−an+11)⇒d1[(a11−a21)+(a21−a31)+….(a201−a211)]=94
⇒d1(a11−a211)=94
⇒d1(a1⋅a21a21−a1)=94⇒d(a1a21)20d=94⇒a1a21=45
a1(a1+20d)=45...(1)
Now sum of first 21 terms=221(2a1+20d)=189⇒a1+10d=9...(2)
by using equation (1) and (2) we get a1=3,d=53 otherwise a1=15,d=−53
So, a6a16=(a1+5d)(a1++15d)=72