Arithmetic sequence or geometric sequence?

The terms of the sequence are the sum of two consecutive cubes, i.e. $-1^{3} + 0^{3} , 0^{3} + 1^{3} , 1^{3} + 2^{3} , 2^{3} + 3^{3} ,3^{3} + 4^{3} ....$ so the next term is $4^{3} + 5^{3} = 189$ .

OEIS has actually two solutions to this question: 189 and 242. 189 is what you said; but 242 is made like this:

Suppose you have the list of whole numbers starting with 0: 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,...

Group them like this: {0,1},{2,3,4},{5,6,7,8,9},{10,11,12,13,14,15,16,17},{18,19,20,21,22,23,24,25,26,27,28},{29,30,31,32,33,34,35,36,37,38,39,40}...

The first set has 2 numbers. The second set has 3 numbers. The third set has 5 numbers. The fourth set has 7 numbers. The fifth set has 11 numbers. The sixth set has 13 numbers. ... The n-th set has $p_n$ numbers, while $p_n$ denotes " the n-th prime number ".

Add all the elements up, you will get: $S_1 = 1, S_2 = 9, S_3 = 35, S_4 = 91, S_5 = 242, S_6 = 442 ...$ While $S_n$ denotes " the sum of all numbers in the n-th set " So you will get a sequence: 1,9,35,91,242,442,... (A034957)