3;3;9;6;15;12;...

1. Writing next TWO terms, by analysis:

The first term and third term(3;9...) onwards of combined sequence form an arithmetic sequence with d=6 (T2-T1).

Second term and fourth term(3;6...) onwards of combined sequence form a geometric sequence with r = 2 (T2/T1).

Therefore next 2 terms are (15+6);(12x2);... = 21;24;...

2. Calculating T52 - T51:

Arithmetic seq terms will take odd number positions (3;9;...= T1;T3;...)

Consequently geometric seq terms take even number positions(T2;T4;...)

Hence T52 is under geometric terms and T51 is under arithmetic terms.

T52 will be T26 (52/2) on its geometric seq and T51 will be therefore before T26, and be T25.

T52-T51 = T26 - T25 = [ar^(n-1) ] - [a + (n-1)d] = [3(2)^(26-1)] - [3 + (25-1)6] = 100 663 296 - 147 = 100 663 149

3. ALL the terms of this infinite sequence to be divisible by 3:

Terms of arithmetic seq (3;9;...) start with a multiple of 3 and d = 6 which is also an additive multiple of 3.

Terms of geometric seq (3;6;...) start also with a multiple of 3 and r = 2 which doubles the multiple and still retains divisibility by 3.

Therefore all terms of the combined seq will be divisible by 3.