The first three terms of an arithmetic sequence are given by $2x + 5$ $5x + 1$ $6x + 7$ 9 (a) Show that $x = 5$ is the only value which gives an arithmetic sequence - AQA - A-Level Maths Pure - Question 9 - 2022 - Paper 1

The first term of the sequence can be calculated by substituting $x = 5$ into the expression for the first term:

$T_1 = 2x + 5 = 2(5) + 5 = 10 + 5 = 15.$

Thus, the first term is $15$.

To find the common difference, we can use the differences calculated from the terms when $x = 5$:

$T_2 - T_1 = 26 - 15 = 11.$

Thus, the common difference of the sequence is $11$.

The sum of the first $N$ terms of an arithmetic sequence can be calculated using the formula:

$S_N = \frac{N}{2} (T_1 + T_N)$

Where $T_N$ is the last term:

1. Substituting for $T_1 = 15$ and the formula for $T_N = T_1 + (N-1)d$ gives us:

   $S_N = \frac{N}{2} (15 + (15 + (N-1) imes 11)) = \frac{N}{2}(15 + 15 + 11N - 11) = \frac{N}{2}(11N + 19)$

2. We need to find $N$ such that:

   $\frac{N(11N + 19)}{2} < 100,000$

   Simplifying yields:

   $N(11N + 19) < 200,000$

   Using trial and improvement or substituting, we find:

   $N = 133$ satisfies $S_{133} < 100,000$. For $N = 134$, we check and find $S_{134} > 100,000$.

Thus, the value of $N$ is $133$.