Calculating the nth Term of Sequences for the TMUA

If we have a list of terms in a sequence, we can determine the $n^{th}$ term, which is the position to term rule used to calculate any term based on its position $n$.

Finding the nth Term for a Linear Sequence

In a linear sequence, the terms increase or decrease by the same amount each time. This is known as a constant difference. To find the $n^{th}$ term, identify this difference and adjust the formula to match the sequence.

Example: Increasing Linear Sequence

Find the $n^{th}$ term for the sequence $2, 5, 8, 11, ...$

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

A constant difference of $+3$ means the $n^{th}$ term rule includes $3n$. Comparing $3n$ (which produces $3, 6, 9, 12$) to our sequence, we see each term is $1$ less than $3n$. Therefore, the $n^{th}$ term is $3n - 1$.

Example: Decreasing Linear Sequence

Find the $n^{th}$ term for the sequence $14, 8, 2, -4, ...$

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

The difference between terms is $-6$. This indicates the $n^{th}$ term involves $-6n$. To get from the $-6n$ values to the actual terms, we must add $20$. Thus, the $n^{th}$ term is $-6n + 20$.

Example: Linear Sequence with Fractional Coefficient

Find the $n^{th}$ term for the sequence $5\frac{1}{2}, 6, 6\frac{1}{2}, 7, ...$

$n$	$1$	$2$	$3$	$4$
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 rule includes $\frac{1}{2}n$. Since each term is $5$ more than $\frac{1}{2}n$, the $n^{th}$ term is $\frac{1}{2}n + 5$.

Finding the nth Term for a Quadratic Sequence

For a quadratic sequence, the first differences are not constant, but the second differences are. The rule is in the form $an^2 + bn + c$.

Method 1: Using Simultaneous Equations

Find the $n^{th}$ term for $2, 5, 10, 17, 26, 37, ...$

1. Calculate the differences:
   • First differences: $+3, +5, +7, +9, +11$
   • Second differences: $+2, +2, +2, +2$
2. Set up equations using $an^2 + bn + c$:
   • When $n = 1: a + b + c = 2$ (term 1)
   • When $n = 2: 4a + 2b + c = 5$ (term 2)
   • When $n = 3: 9a + 3b + c = 10$ (term 3)
3. Subtract equations:
   • $(4a + 2b + c) - (a + b + c) \implies 3a + b = 3$
   • $(9a + 3b + c) - (4a + 2b + c) \implies 5a + b = 5$
4. Solve for $a$ and $b$:
   • $(5a + b) - (3a + b) \implies 2a = 2$, so $a = 1$
   • Substitute $a=1$ into $3a + b = 3 \implies 3(1) + b = 3$, so $b = 0$
   • Substitute $a=1, b=0$ into $a + b + c = 2 \implies 1 + 0 + c = 2$, so $c = 1$

The $n^{th}$ term is $n^2 + 1$.

Method 2: Comparing with $n^2$

Using the same sequence ($2, 5, 10, 17...$), the second difference is constant ($+2$). The coefficient $a$ is half the second difference: $a = 2 \div 2 = 1$. The rule involves $1n^2$. Subtract $n^2$ values ($1, 4, 9, 16$) from the original terms to find the remaining linear sequence:

• $2 - 1 = 1$
• $5 - 4 = 1$
• $10 - 9 = 1$ Since the remainder is always $1$, the rule is $n^2 + 1$.

Finding the nth Term for a Quadratic Based on a Multiple of $n^2$

Find the $n^{th}$ term for $1, 10, 25, 46, ...$

• First differences: $+9, +15, +21$
• Second differences: $+6, +6$ Since the second difference is $+6$, $a = 6 \div 2 = 3$. The rule involves $3n^2$. The values for $3n^2$ are $3, 12, 27, 48$. Comparing these to $1, 10, 25, 46$, we see each term is $2$ smaller. The $n^{th}$ term is $3n^2 - 2$.

Quadratic Sequences with $n^2$ and $n$ terms

Find the $n^{th}$ term for $5, 14, 27, 44, 65, ...$

• First differences: $+9, +13, +17, +21$
• Second differences: $+4, +4, +4$ Coefficient $a = 4 \div 2 = 2$. The rule involves $2n^2$. Subtract $2n^2$ from the original terms:
• $5 - 2(1)^2 = 3$
• $14 - 2(2)^2 = 6$
• $27 - 2(3)^2 = 9$ The remainder sequence $3, 6, 9, ...$ is a linear sequence with $n^{th}$ term $3n$. Combining these, the final rule is $2n^2 + 3n$.