Please help me with this Arithmetic Progression question!

The 3rd and 6th terms of an A.P. are 21 and 15 respectively. Find the first term and the common difference of the progression, and hence, find the first term that has a negative value.

Here are the terms:
1st … a(1)
2nd … a(1) + d
3rd … a(1) + 2d = 21
4th … a(1) + 3d
5th … a(1) + 4d
6th … a(1) + 5d = 15

3d = 15 - 21 = -6
d = -2

a(1) + 2(-2) = 21
a(1) - 4 = 21
a(1) = 25

nth term = 27 - 2n
27 - 2n < 0
-2n < -27
n < -13.5
n > 13.5

The 14th term is the first negative one. (Its value = -1)

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a(3) = 3d + c = 21

a(6) = 6d + c = 15

-3d = 6 --> d = -2

c = 27

a(n) = -2n + 27

a(1) = 25

a(n) = -2n + 27 < 0

n > 27/2

n = 14

15 = 21 +3d ---> d = -2
21 = a + 2d ---> a = 25

t_n = 25 - 2(n-1) = 27 - 2n < 0 ---> n = 14