If 4 times the 4th term of an AP is equal to 18 times its 18th term then find its 22nd term.

Let a be the first term and d be the common difference of the AP. Then,

$4 \times a_{4}=18 \times a_{18} \quad$ (Given)

$\Rightarrow 4(a+3 d)=18(a+17 d) \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow 2(a+3 d)=9(a+17 d)$

$\Rightarrow 2 a+6 d=9 a+153 d$

$\Rightarrow 7 a=-147 d$

$\Rightarrow a=-21 d$

$\Rightarrow a+21 d=0$

$\Rightarrow a+(22-1) d=0$

$\Rightarrow a_{22}=0$

Hence, the 22nd term of the AP is 0.