The n th term of an arithmetic sequence is given by the following expression, for n ∈ N: T_n = -254 + (n - 1)(4) (a) (i) Find the value of T_1, the first term of this sequence - Leaving Cert Mathematics - Question 6 - 2022

First, calculate T_2 by substituting n = 2 into the formula:

$$T_2 = -254 + (2 - 1)(4) = -254 + 4 = -250$$

Now, find the common difference:

$$T_2 - T_1 = -250 - (-254) = -250 + 254 = 4$$

Therefore, the common difference is 4.

Solve the inequality:

$$-254 + (n - 1)(4) > 0$$

First, simplify the inequality:

$$(n - 1)(4) > 254$$

Dividing both sides by 4 gives:

$$n - 1 > 63.5$$

Thus,

$$n > 64.5$$

The smallest integer value satisfying this is n = 65.

Set up the equation:

$$\frac{n}{2} [2(-254) + 4n - 4] = 0$$

This implies:

$$\frac{n}{2} [-508 + 4n - 4] = 0$$

Simplifying further:

$$\frac{n}{2} [4n - 512] = 0$$

This leads to two potential solutions:

1. \frac{n}{2} = 0, which cannot be as n ≠ 0.
2. 4n - 512 = 0

From this, we solve:

$$4n = 512 \ n = \frac{512}{4} = 128$$

Thus, the value of n is 128.