Let a be the first term and d be the common difference of A.P.
Given, S7S11=611
⇒7(2a+6d)11(2a+10d)=611⇒7(a+3d)=6(a+5d)⇒a=9d ⋯(1)
Now, we know that,
130<a7<140⇒130<a+6d<140
Using equation (1), we get
130<15d<140⇒263<d<283
As given A.P. consists of natural numbers, so d=9