Given: a₂ = 38 and a₆ = –22

We know that,

a_n = a + (n – 1)d

a₂ = a + (2 – 1)d

38 = a + d …(i)

and a₆ = a +(6 – 1)d

–22 = a + 5d …(ii)

Solving linear equations (i) and (ii), we get

a + d – a – 5d = 38 – (–22)

⇒ –4d = 60

⇒ d = –15

Putting the value of d in eq. (i), we get

a + (–15) = 38

⇒ a – 15 = 38

⇒ a = 53

Now, a₃ = a + 2d = 53 + 2(–15) = 53 – 30 = 23

a₄ = a + 3d = 53 + 3(–15) = 53 – 45 = 8

a₅ = a + 4d = 53 + 4(–15) = 53 – 60 = –7

Hence, the missing terms are 53, 23, 8 and –7