QUESTION 3 The following sequence is a combination of an arithmetic and a geometric sequence: \[ 3: 3: 9: 6: 15: 12: \ldots \] 3.1 Write down the next TWO terms. 3.2 Calculate \( T_{52}-T_{34} \). 3.3 Prove that ALL the terms of this infinite sequence will be divisible by 3 .

To find the next two terms in the sequence, we can observe how the terms alternate between being multiplied and added consistently. Starting from your sequence: - The first term (3) stays the same, then the next is also 3, then it follows the pattern of multiplying by 3 for 9, then subtracting 3 for 6, adding 9 (3*3) for 15, and subtracting 3 for 12. - Continuing this pattern, the next term after 12 will be \(3 + 3 \times 3 = 3 + 9 = 21\) and the next term after that will be \(12 + 3 = 27\). So the next two terms are: **21, 27**. For \( T_{52} - T_{34} \): To calculate \( T_{52} \) and \( T_{34} \), we can use the established pattern. Since the sequence alternates, we can express \( T_n \) based on whether \( n \) is odd or even. - Odd positions (like 1, 3, 5...) represent where the sequence behaves like \(3 + (n-1) \times 3\). - Even positions (like 2, 4, 6...) are based directly on the multiplication. By establishing direct formulas, we can find that after consistent application: - \( T_{52} = 3 + (52-1)/2 \times 6 = 3 + 25 \times 6 = 3 + 150 = 153\) - \( T_{34} = 3 + (34-1)/2 \times 6 = 3 + 16 \times 6 = 3 + 96 = 99 \) Thus, \( T_{52} - T_{34} = 153 - 99 = 54 \). Lastly, to prove that all terms are divisible by 3: Every term in the sequence can be expressed in the form of either \( 3 + 3k \) or \( 6m \) (where k and m are integers), which inherently ensures their divisibility by 3 due to the presence of the factor 3. Thus, every term will indeed always be divisible by 3.