Correct option is (B) 0
Given that 7 times of \(7^{th}\) term is equal to 11 times of \(11^{th}\) term.
i.e., \(7a_7=11a_{11}\)
\(\Rightarrow7(a+6d)=11(a+10d)\) \((\because a_n=a+(n-1)d)\)
\(\Rightarrow11a+110d-7a-42d=0\)
\(\Rightarrow4a+68d=0\)
\(\Rightarrow4(a+17d)=0\)
\(\Rightarrow a+17d=0\)
\(\Rightarrow a_{18}=0\) \((\because a_{18}=a+(18-1)d=a+17d)\)
Hence, 18th term of given A.P. is 0.
