Question

find the sum of the terms of an a.p. 2,6,0,114,....,32

Answer 1

This question is wrong. The terms of the given AP must be like this..
2, 6, 10, 14, 18, 22, 26, 34 .... but there is no 32. I think the total terms can be 32. if it is so then the solution given by me is correct otherwise not. 
Solution:-
a = 2 and d = 4
Sn = n/2 {2a + (n - 1)d}
Sn = 32/2{2*2 + (32 - 1)4}
Sn = 16(4 + 31*4)
Sn = 16*128
Sn = 2048
So, the sum of the given AP is 2048

Answer 2

I assume 32 is the number of terms
than
first term= a= 2 
common difference= d= 4

Sn = n/2 [2a + (n - 1)d]
Sn = 32/2 [2x2+ (32 - 1)4]
Sn = 16 (4 + 31 x 4)
Sn = 16 x 128
Sn = 2048