Arithmetic progression

Please help me to solve this problem.

In arithmetic sequence, the sum of the first twelve terms is 156 and the 4th term is 8. Find
a) the first term and the common difference,
b) the tenth term

This is my solution for (a) ,

156= 12/2( 2a + (n-1) )
156= 6 ( 2a+11d )
156= 12a + 66d equation (1)

8= a + 3d equation (2)
a= 8 - 3d

Then I substituted the equation

156= 12( 8 - 3d) + 66d
156= 96-36d + 66d
60= 30d
2=d but the answer for d is 3

I substituted d into a to find the first term

a= 8 - 3d
a= 8 - 3(2)
a= 2 but the answer for a is -1

Can anyone help to to solve this problem, I need the answer as soon as possible.
Thanks for helping me.

S12=156
=>12/2(2a+(12-1)d)=156
6(2a+11d)=156
or
2a+11d=26---------------(i)
similarly for a4=8 we have
a+3d=8------------------(ii )
a=8-3d----------------(iii)
using in (i)
16-6d+11d=26
16+5d=26
5d=26-16
5d=10
d=2
a=8-6=2
a10=a+9d
a10=2+9(2)
a10=20
your answer seems correct to me

In arithmetic sequence, the sum of the first twelve terms is 156 and the 4th term is 8. Find
a) the first term and the common difference,
b) the tenth term
ANSWER:
AS be (a, d) then 6(2a + 11d) = 156 and a + 3d = 8
whence substituting a = (8 -- 3d) from second eqn into first eqn
6(16 -- 6d + 11d) = 156 whence d = 2 and a = 2 Answers
hence tenth term = a + 9d = 2 + 9*2 = 20 Answer.
Answer for a = -- 1 IS WRONG.

2+4+6+8 +...+24 =156
a=2, d=2
but -1 +2+5 +8 + ... + 32 =186
where a=-1, d=3
so the way you did it is correct, the answer is wrong