The 9th term of an A.P. is equal to 6 times its second term.

Let a be the first term and d be the common difference.

We know that, nth term = an = a + (n − 1)d

According to the question,

a9 = 6a2
⇒ a + (9 − 1)d =  6(a + (2 − 1)d)
⇒ a + 8d =  6a + 6d
⇒ 8d − 6d =  6a − a
⇒ 2d = 5a

$\Rightarrow a=\frac{2}{5} d$

Also, a5 = 22
⇒ a + (5 − 1)d = 22
⇒ a + 4d = 22   ....(2)

On substituting the values of (1) in (2), we get

$\frac{2}{5} d+4 d=22$

$\Rightarrow 2 d+20 d=22 \times 5$

$\Rightarrow 22 d=110$

⇒ d = 5

$\Rightarrow a=\frac{2}{5} \times 5 \quad[$ From (1) $]$

⇒ a = 2

Thus, the A.P. is 2, 7, 12, 17, .... .