Answer 1

5, 1, – 3, …

d = t₂ – t₁ = t₃ – t₂

⇒ 1 – 5 = – 3 – 1

– 4 = – 4

∴ It is an A.P.

t_n = a + (n – 1)d

t₄ = 5 + 3 × – 4

= 5 – 12

= – 7

15 = a + 4d

= 5 + 4 × – 4

= 5 – 16

= – 11

t₆ = a + 5d

= 5 + 5 × – 4

= 5 – 20

= – 15

∴ The next three terms are 5, 1, – 3, – 7, – 11, – 15.