Answer:
[tex][/tex] To prove that the sequence defined by Un = 3n - 2 is arithmetic, we need to show that the difference between consecutive terms is constant. Let's find the difference between Un+1 and Un.
Un+1 = 3(n+1) - 2 = 3n + 3 - 2 = 3n + 1
Un = 3n - 2
Now, we calculate Un+1 - Un:
Un+1 - Un = (3n + 1) - (3n - 2)
Un+1 - Un = 3n + 1 - 3n + 2 We subtract the terms in the brackets
Un+1 - Un = 3n - 3n + 1 + 2
Un+1 - Un = 1 + 2 Solving the addition
Un+1 - Un = 3
Since Un+1 - Un is a constant value (3), the sequence is arithmetic.
Now, to find the first term (U1) and the common difference (d) in an arithmetic sequence, we look at the formula for an arithmetic sequence:
Un = U1 + (n-1)d
Given that Un = 3n - 2 and that the sequence is arithmetic with a common difference of 3, we can set up the following equation:
3n - 2 = U1 + (n-1)3
Let's solve for U1 and d:
U1 + (n-1)3 = 3n - 2
U1 + 3n - 3 = 3n - 2
U1 - 3 = -2
Combine like terms:
U1 = 3 - 2
U1 = 1
Therefore, the first term U1 of the arithmetic sequence is 1, and the common difference d is 3.