91Ó°ÊÓ

In an arithmetic sequence, the common difference, often denoted by the letter \(d\), is the amount added to each term to get the next one. It's like the step size of the sequence, helping us find any term if we have enough information. If the sequence starts with \(a_1\), the terms will progress as follows:

• \(a_2 = a_1 + d\)
• \(a_3 = a_2 + d\)
• And so forth...

The common difference is constant and can be positive, negative, or zero, indicating the direction and rate of change of the sequence. In our example, the common difference is given as \(d = 9\), meaning each term is 9 more than the previous one. This regular increment is crucial because it helps establish a pattern and predict future or past terms.

In certain problems involving arithmetic sequences, we encounter quadratic equations. These are versatile and arise here when solving for the number of terms \(n\). The general form of a quadratic equation is \(ax^2 + bx + c = 0\).
The solution to our problem led to the quadratic equation \(9n^2 - 181n + 910 = 0\). This form came from manipulating our equations to incorporate both \(n\) and known variables. Quadratics like this one are typically solved using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Where \(a = 9\), \(b = -181\), and \(c = 910\) in our case.
Solving this gives us the potential number of terms in our sequence, showing how quadratics play a role in tying sequence values together logically.

The sum formula for arithmetic sequences is a powerful tool for finding the sum of a certain number of terms. This formula states that:
\[ S_n = \frac{n}{2} (a_1 + a_n) \]This means the sum of the first \(n\) terms is the average of the first and last term, multiplied by the number of terms. In simpler words, it’s like finding the midpoint between two ends, then multiplying by how many steps there are.
Using our problem, we had:\[ 455 = \frac{n}{2} (a_1 + 86) \]Given \(S_n = 455\), the sum formula helped relate these values to solve for unknowns by balancing the equation. The sum formula works as a bridge, connecting known and unknown elements of the sequence.

The last term formula allows us to express any specific term of the sequence in relation to the first term and the common difference. It is given by the formula:
\[ a_n = a_1 + (n-1) \times d \]This formula tells us how to find the nth term, where the expression \((n-1)\times d\) accounts for the sequence progression times the step size \(d\).
In our given problem, \(a_n = 86\) enabled forming:\[ 86 = a_1 + (n-1) \times 9 \]The equation highlights that knowing the last or any term, along with how far along it is in the sequence, connects back to \(a_1\) and \(d\). It aids in completing the picture of the sequence dynamics.