Find the sum of the first 12th terms of the sequence 2,5,8,11

Answer 1

term n = a + d(n-1)
a = 2
d = 3

sum from 0 to n = (n/2) [ 2 a +(n-1)d ]
n = 12
sum = 6 [ 4 + 11*3] = 6 * 37

Answer 2

Is far final answer

Answer 3

222

Answer 4

T12 =s=n/2 [2a+(n-1)d
a=2
d=5-2=3
S=12/2[2*2+(12-1)3]
[4+11*3]
[6*37]
= 222

Answer 5

227

Answer 6

227

Answer 7

I apologize, but I couldn't quite understand the context of this response. Could you please provide more information or clarify your question?

Answer 8

an=a1+(n-1)
a1=2
d=3
an=2+(n-1)×3
=2+3n-3
=3n-3
S=(n/2) (a1+an)
S=(12/2) [2+(3×12-1)]
6(2+35)
6=37
=222