If 9 times the 9th term of an AP is equal to 13 times the 13th term,

Given : $9 \times\left(9^{\text {th }}\right.$ term $)=13 \times\left(13^{\text {th }}\right.$ term $)$

To prove: $22^{\text {nd }}$ term is 0

$9 \times(a+8 d)=13 \times(a+12 d)$

$9 a+72 d=13 a+156 d$

$-4 a=84 d$

$a=-21 d \ldots . .$ Equation 1

Also $22^{\text {nd }}$ term is given by

a + 21d

Using equation 1 we get

$-21 d+21 d=0$

Hence proved 22nd term is 0.