Arithmetic sequence問題 急!!!

1 )
Consideran arithmetic sequence with the fourth term is 53 and the second term is threetimes the 21st term .
a) Find the first Term and the commondifference.
b) Find the sum of the first 21terms of the arithmetic sequence

Solution :
a)
First term, T(1) = a
Common difference = d

T(4) = 53
a + 3d = 53
a = 53 - 3d ...... [1]

T(2) = 3T(21)
a + d = 3(a + 20d)
a + d = 3a + 60d
2a + 59d = 0 ...... [2]

Put [1] into [2] :
2(53 - 3d) + 59d = 0
106 - 6d + 59d = 0
53d = -106
d = -2

Put d = -2 into [1] :
a = 53 - 3*(-2)
a = 59

Hence, the first term is 59 and the common difference is -2.

b)
Sum of the first 21st term
= n[2a + (n - 1)d/2
= 21[2*59 + 20*(-2)]/2
= 819


2 )
The sum of the first n termsof an arithmetic sequence is n² + 6n. find the nth term of the sequence.

Solution :
First term, T(1) = a
Common difference = d

S(n) = n² + 6n
n[2a + (n - 1)d]/2 = n² + 6n
2a + (n - 1)d = 2n + 12
2a + nd - d = 2n + 12
dn + (2a - d) = 2n + 12

Compare the both sides:
d = 2 and 2a - d = 12
d = 2 and 2a - 2 = 14
d = 2 and a = 7

The nth term
= a + (n - 1)d
= 7 + (n - 1)*2
= 7 + 2n - 2
= 2n + 5