Linear and Quadratic Sequences for the ESAT

When we are presented with a list of terms in a sequence, we can determine the $n^{\text{th}}$ term. This rule allows us to calculate any value in the sequence simply by knowing its position. For example, in the linear sequence $1, 3, 5, 7, \dots$, each term increases by $2$ every time, resulting in an $n^{\text{th}}$ term of $2n - 1$. In the quadratic sequence $1, 7, 17, 31, \dots$, the values are generated using the rule $2n^2 - 1$.

Finding the $n^{\text{th}}$ term for a linear sequence

A linear sequence is defined by terms that increase or decrease by the same amount each time, meaning there is a constant difference between them. Consider the sequence $2, 5, 8, 11, \dots$:

$n$	1	2	3	4
term	2	5	8	11
difference		+3	+3	+3

Because there is a constant difference of $+3$, the $n^{\text{th}}$ term must include $3n$. We observe that each term in the sequence is exactly $1$ less than $3 \times n$ (for example, when $n=1$, $3(1) = 3$ and $3 - 1 = 2$). Therefore, the $n^{\text{th}}$ term is $3n - 1$.

Finding the $n^{\text{th}}$ term for a decreasing linear sequence

For sequences where the values decrease, the method remains the same but involves negative coefficients. Consider the sequence $14, 8, 2, -4, \dots$. We can organise this by writing the position numbers $n$ and the terms, leaving space for a middle row to help our calculation:

$n$	1	2	3	4
$-6n$	-6	-12	-18	-24
term	14	8	2	-4

Since the constant difference between terms is $-6$, we know the $n^{\text{th}}$ term involves $-6n$. Comparing the values of $-6n$ to the actual terms in the sequence, we see that we must add $20$ to each middle row value to reach the term (for example, $-6 + 20 = 14$). Thus, the $n^{\text{th}}$ term is $-6n + 20$.

Finding the $n^{\text{th}}$ term for a linear sequence with fractional coefficient

Linear sequences can also have differences that are not integers. Consider the sequence $5\frac{1}{2}, 6, 6\frac{1}{2}, 7, \dots$:

$n$	1	2	3	4
term	$5\frac{1}{2}$	6	$6\frac{1}{2}$	7
difference		$+\frac{1}{2}$	$+\frac{1}{2}$	$+\frac{1}{2}$

The constant difference is $+\frac{1}{2}$, so the $n^{\text{th}}$ term includes $\frac{1}{2}n$. By comparing $\frac{1}{2}n$ to the terms, we see that each term is $5$ more than its corresponding $\frac{1}{2}n$ value (for $n=1$, $\frac{1}{2} + 5 = 5\frac{1}{2}$). The $n^{\text{th}}$ term is $\frac{1}{2}n + 5$.

Finding the $n^{\text{th}}$ term for a quadratic sequence

A quadratic sequence is one where the first differences change, but the second differences (the difference between the differences) are constant. The general form is $an^2 + bn + c$. Consider the sequence $2, 5, 10, 17, 26, 37, \dots$.

$n$	1	2	3	4	5	6
term	2	5	10	17	26	37
1st difference		+3	+5	+7	+9	+11
2nd difference			+2	+2	+2	+2

Method 1: Simultaneous Equations We know the form is $an^2 + bn + c$. We can create equations using values of $n$:

1. When $n=1$, $a + b + c = 2$ (the 1st term).
2. When $n=2$, $4a + 2b + c = 5$ (the 2nd term).
3. When $n=3$, $9a + 3b + c = 10$ (the 3rd term).

Subtracting equation 1 from equation 2 gives $3a + b = 3$. Subtracting equation 2 from equation 3 gives $5a + b = 5$. Solving these two new equations: $(5a + b) - (3a + b) = 5 - 3$, so $2a = 2$, which means $a = 1$. Substituting $a=1$ into $3a + b = 3$ gives $b = 0$. Finally, substituting into $a + b + c = 2$ gives $1 + 0 + c = 2$, so $c = 1$. The $n^{\text{th}}$ term is $n^2 + 1$.

Method 2: Comparing to $n^2$ To find the coefficient $a$, divide the second difference by two: $2 \div 2 = 1$. This tells us the rule involves $1n^2$. The sequence for $n^2$ is $1, 4, 9, 16$. Comparing these to our sequence $2, 5, 10, 17$, we see every term is $1$ greater than $n^2$. Thus, the rule is $n^2 + 1$.

Finding the $n^{\text{th}}$ term for a quadratic sequence based on a multiple of $n^2$

Consider the sequence $1, 10, 25, 46, \dots$. 1st differences: $+9, +15, +21$. Second difference: $+6$. Using Method 1, we establish $a+b+c=1$ and $4a+2b+c=10$, giving $3a+b=9$. With $n=3$, $9a+3b+c=25$, so $5a+b=15$. Solving these, $2a=6$ hence $a=3$, $b=0$, and $c=-2$. The rule is $3n^2 - 2$. Using Method 2, $a = 6 \div 2 = 3$. The sequence $3n^2$ would be $3, 12, 27, 48, \dots$. Each term in our actual sequence is $2$ smaller than these values, so the rule is $3n^2 - 2$.

Finding the $n^{\text{th}}$ term for a quadratic sequence with terms in $n^2$ and $n$

Consider the sequence $5, 14, 27, 44, 65, \dots$. 1st differences: $+9, +13, +17, +21$. Second difference: $+4$. Method 1: Using $n=1, 2, 3$, we get $a+b+c=5$, $4a+2b+c=14$, and $9a+3b+c=27$. Subtracting gives $3a+b=9$ and $5a+b=13$. Solving these gives $2a=4$, so $a=2$. Then $3(2)+b=9$, so $b=3$. Finally, $2+3+c=5$, so $c=0$. The rule is $2n^2 + 3n$.

Method 2: Since $a = 4 \div 2 = 2$, the term involves $2n^2$. Compare the sequence to $2n^2$:

term	5	14	27	44	65
$2n^2$	2	8	18	32	50
difference	+3	+6	+9	+12	+15

The differences $+3, +6, +9, +12, +15$ form a linear sequence with $n^{\text{th}}$ term $3n$. Combining these parts, the full rule is $2n^2 + 3n$.