sn = u + a/2 (2n -1)
s2 = 12
⇒ 12 = u+ a/2 (2 × 2 – 1)
⇒ 12 = u + 3 a/2 ………. (1)
s4 = 20
⇒ 20 = u + a/2 (2 × 4 – 1)
⇒ 20 = u + 7a/2 ………. (2)
Subtract equation (1) from (2)
s = 4a/2
⇒ a = 4 ms-2
12 = u + 3/2 × 4
⇒ u = 6 ms-1
Distance covered in the 4 seconds after 5th second,
= S9 – S5
= u (9) + 1/2 a (9)2 – [u(5) + 1/2 a (5)2]
= 4u + a/2 (81 – 25)
= 4 × 6 + 4/2 × 56 = 136 m.