Consider the quadratic number pattern: -145; -122; -101; .. - NSC Mathematics - Question 3 - 2021 - Paper 1

The general term is derived from the differences:

1. The first difference is $23$ and the second difference is $21$, indicating the presence of a quadratic relationship.
2. Setting $a = -2$ and $b = 26$ gives us the equation:

$T_n = -n^2 + 26n - 170.$

To find which two terms have a difference of -121, we solve:

$T_n - T_{n-1} = -121.$ Using the general formula, we set this up:

$(-n^2 + 26n - 170) - (-(n-1)^2 + 26(n-1) - 170) = -121.$ After simplifying and solving for $n$, we find:

Between $T_{3}$ and $T_{4}.$

The maximum term in the original pattern occurs when $n$ is at its peak before becoming negative. We can find this using:

T_{n_{max}} = -rac{b}{2a}. Substituting $a = -1$ and $b = 26$, we find:

n = rac{-26}{2(-1)} = 13. Thus, the maximum term:

$T_{13} = - (13^2) + 26(13) - 170 = -169 + 338 - 170 = -1.$ To make this equal to 1, we need:

$1 - (-1) = 2.$ Thus, we need to add $2$ to each term.