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

Given: Sum of first m terms of an AP is given by:

S_m = [2a + (m - 1)d] = 4m² - m

Now, n^th term is given by: a_n = S_n - S_{n - 1}

∴ a_n = (4n² - n) - [4(n - 1)² - (n - 1)]

= (4n² - n) - [4(n² + 1 - 2n) - n + 1]

= 4n² - n - 4n² - 4 + 8n + n - 1

= 8n - 5 …………………(1)

Now, given that a_n = 107

⇒ 8n - 5 = 107

⇒ 8n = 112

⇒ n = 14

For 21^st term of AP, put n = 21 in the value of the nth term in equation (1), we get

a₂₁ = 8 × (21) - 5

⇒ a₂₁ = 168 - 5

= 163

∴ a₂₁ = 163