Given, first term a = 2

Let d be the common difference

Let S_n denote sum of n terms,

t_n denote n^th term.

t_n = a + (n – 1)d

S₁ = Sum of first five terms.

S₂ = Sum of next five terms.

S₁ = 10 × (1 + d)

S₂ = 20 + 45d – 10 – 10d = 10 + 35d

Also, S₁ = 1/4 S₂

⇒ 10 + 10d = 1/4 (10 + 35d)

⇒ 40 + 40d = 10 + 35d

⇒ 30 = -5d

⇒ d = -6

t₂₀ = 2 + (20 – 1) × (-6)

t₂₀ = 2 – 19 × 6 = 2 – 114

t₂₀ = -112