If 7 times the 7th term of an AP

(d) According to the question,

$7 a_{7}=11 a_{11}$

$\Rightarrow \quad 7[a+(7-1) d]=11[a+(11-1) d] \quad\left[\because a_{n}=a+(n-1) d\right]$

$\Rightarrow \quad 7(a+6 d)=11(a+10 d)$

$\Rightarrow \quad 7 a+42 d=11 a+110 d$

$\Rightarrow \quad 4 a+68 d=0$

$\Rightarrow \quad 2(2 a+34 d)=0$

$\Rightarrow \quad 2 a+34 d=0$  $[\because 2 \neq 0]$

$\Rightarrow \quad a+17 d=0$ ...(i)

$\therefore \quad$ 18th term of an AP, $a_{18}=a+(18-1) d$

$=a+17 d=0 \quad$ [from Eq. (i)]