Finding the nth Term of a Linear Sequence

Substitute $n=5$ into the rule: $6\times 5-1=30-1=29$.

So the 5th term is 29.

The rule has the form $an+b$.

The common difference is $a=8-6=2$.

The adjustment is $b=\text{first term}-a=6-2=4$, so the nth term is $2n+4$.

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

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

Then $b=4-7=-3$, so the nth term is $7n-3$.

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

So the 20th term is 137.

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

The adjustment is $b=30-(-4)=34$, 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.

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

Solving gives $2n=36$, so $n=18$.

Since 18 is a positive whole number, 41 is in the sequence. It is the 18th term.

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 $n=14.57$.

Since $n$ is not a positive whole number, 100 is not in the sequence. You cannot round $n$ to a whole number, so there is no '14.57th term'.