a body travels 2 m in 2^nd second & 6 m in next 4 seconds. what will be the distance travelled in 9^th second?

For nth second the distance travelled is given as--
S = u + a/2 (2n + 1)
where,
S = distance travelled
u = initial velocity
a = acceleration
For 2^nd second,
2 = u + a/2 (2* 2 + 1) ------ (as distance travelled in 2^nd sec is 2m)
2u + 5a = 4 -------- (1)
Now for distance travelled in 4^th second,
S = ut + ½ at²
6 = 4u + ½ a (4)² --------- (distance travelled in 4sec is 6m)
2u + 4a = 3 --------(2)
On solving (1) and (2) we get,
a = 1m/s² and putting this value in any equation we get, u = – 0.2m/s( negative sign indicates the opposite direction)
thus for 9^th second,
S = u + a/2 (2n+1)
S = -0.5 + 1/2(2*9 + 1)
S = 9m