Let a = 2 be the first term and d be the common difference. ∴ sum of first five terms $S_5$ = $\frac{5}{2}$ [4 + (5 - 1) d] = 5 (2 + 2d) ... (i) sum of first ten terms $S_{10}$ = $\frac{10}{2}$ [4 + (10 - 1) d] = 5 (4 + 9d) ... (ii) According to question, Sum of first five terms = $\frac{1}{4}$ (sum of next five terms) ⇒ $S_5$ = $\frac{1}{4}(s_{10}-S_5)$ ⇒ $4S_5$ = $S_{10}-S_5$ ⇒ $5S_5$ = $S_{10}$ ⇒ 5[5(2 + 2d)] = 5(4 + 9d) (From (i) & (ii)) ⇒ 25 (2 + 2d)] = 5 (4 + 9d) ⇒ 50 + 50d = 20 + 45d ⇒ 5d = – 30 ⇒ d = – 6 ∴ $t_{20}$ = a + (20 – 1)d = 2 + (19) (– 6) = 2 – 114 = – 112.