Let S_n denote the sum of the first n terms of an A.P. If S_4=16 and S_6=-48, then S_10 is equal to?

Correct option is -320
2a+3d=16
3(2a+5d)=-48
2a+3d=8
2a+5d=-16
Subtracting the two equations:
-2d=24
d=-12
Substituting d=-12 in 2a+3d=8
2a+3(-12)=8
2a-36=8
2a=44
a=22
S_n = n/2(2a+(n-1)d)
S_10 = 10/2(2(22)+(10-1)(-12))
S_10 = 5(44-108)
S_10 = 5(-64)
S_10 = -320