Let the common difference of an A.P. be d, then
2a7=a5 (given)
⇒2(a1+6d)=a1+4d
⇒a1+8d=0…(1)
And,
a11=18
⇒a1+10d=18…(2)
Solving (1) and (2), we get
a1=−72,d=9
So,
a18=a1+17d=−72+153=81
a10=a1+9d=9
Now,
12(a10+a111+a11+a121+…..a17+a181)
=12(da11−a10+da12−a11+……da18−a17)
=12(da18−a10)=912(9−3)=8