In the given AP, the first term = a = 3

Common difference = d = 8 - 3 = 5

To find: place of the term which is 55 more than its 20^th term.

So, we first find its 20^th term.

Since, we know that

a_n = a + (n - 1) × d

∴ a₂₀ = 3 + (20 - 1) × 5

⇒ a₂₀ = 3 + 19 × 5

⇒ a₂₀ = 3 + 95

⇒ a₂₀ = 98

∴ 20^th term of the AP is 98.

Now, 55 more than 20^th term of the AP is 55 + 98 = 153.

So, to find: place of the term 153.

So, let a_n = 153

Since, we know that

a_n = a + (n - 1) × d

∴ 153 = 3 + (n - 1) × 5

⇒ 153 - 3 = 5n - 5

⇒ 150 = 5n - 5

⇒ 150 + 5 = 5n

⇒ 5n = 155

⇒ n = 155/5 = 31

∴ 31^st term of the AP is the term which is 55 more than 20^th term.