Finding the nth Term of a Linear Sequence

Substitute $n=5$ into the rule: $6n-1=6\times 5-1$.

Multiply first, then subtract: $30-1=29$.

So the 5th term is 29.

Remember $6n$ means $6\times n$ (not $6+n$), and the $-1$ must be done after the multiplication, not before.

The rule has the form $an+b$.

The common difference is $a=8-6=2$, so the rule starts with $2n$, which gives $2,4,6,8,\dots$

Every term of our sequence is 4 more than that, and $b=\text{first term}-a=6-2=4$. So the nth term is $2n+4$.

Check: $n=1$ gives 6 and $n=4$ gives 12.

Note that $b$ is the first term minus $a$, not the first term itself, and not $a$ minus the first term.

First find the rule. The common difference is $a=11-4=7$.

Then $b=\text{first term}-a=4-7=-3$, so the nth term is $7n-3$.

Substitute $n=20$: $7\times 20-3=140-3=137$.

So the 20th term is 137.

Watch the order when finding $b$: it is $4-7=-3$, not $7-4=3$.

The terms are decreasing, so the common difference is $a=26-30=-4$.

Then $b=\text{first term}-a=30-(-4)=34$, so the nth term is $-4n+34$.

Substitute $n=15$: $-4\times 15+34=-60+34=-26$.

So the 15th term is $-26$.

Take care with the signs: subtracting a negative makes $b$ larger, and $-60+34$ comes to $-26$, not 26. The answer stays negative.

Set the rule equal to the number: $2n+5=41$.

Undo the $+5$ first: $2n=41-5=36$. Then undo the $\times 2$: $n=36÷2=18$.

Since $n=18$ is a positive whole number, 41 is in the sequence: it is the 18th term.

Check: $2\times 18+5=36+5=41$.

First find the rule. The common difference is $a=12-5=7$, and $b=5-7=-2$, so the nth term is $7n-2$.

Set it equal to the number: $7n-2=100$, so $7n=102$ and $n=\frac{102}{7}=14.571\dots$

Since $n$ is not a whole number, 100 is not a term in the sequence.

The 14th term is $7\times 14-2=96$ and the 15th is $7\times 15-2=103$, so 100 falls between two terms. You cannot round $n$: there is no '14.57th term'.