GCSE nth term of linear sequences Practice Questions

Find the gaps between consecutive terms

Subtract each term from the one after it.

Check the gaps are all the same

Every gap is 2, so the sequence is linear and the common difference is 2.

State the common difference

The terms increase by the same amount each time.

Find the common difference

The gap between one term and the next is the same every time, so this is a linear (arithmetic) sequence. It goes up in 3s, so the common difference is $d = 3$.

Write down the multiples of the common difference

Because the common difference is 3, the nth term must start with 3n. Working out 3n for $n = 1$, 2, 3, 4 gives 3, 6, 9, 12.

Adjust by the constant and write the nth term

Each term of the sequence is 9 more than the matching multiple of 3, so the nth term is 3n + 9. Check $n = 1$: 3 $\times$ $1 + 9 = 3 + 9 = 12$, the first term.

Find the common difference

The gap between one term and the next is the same every time, so this is a linear (arithmetic) sequence. It goes up in 7s, so the common difference is $d = 7$.

Write down the multiples of the common difference

Because the common difference is 7, the nth term must start with 7n. Working out 7n for $n = 1$, 2, 3, 4 gives 7, 14, 21, 28.

Compare the multiples with the sequence

Every term is 1 more than the matching multiple of 7, so the constant is 1.

Write down the nth term

Putting the 7n and the constant together gives the rule 7n + 1.

Check the rule against the given terms

Substituting $n = 1$ gives 8 and $n = 4$ gives 29, which are the first and last terms given. The rule is correct - always test $n = 1$, because starting the count at the wrong place is the most common mistake.

Use the rule to find the 11th term

Substituting $n = 11$ into 7n + 1 gives 78, so the 11th term is 78. Note that continuing the sequence by adding 7 again and again would take 10 steps - the rule does it in one.

Find the common difference

The gap between one term and the next is the same every time, so this is a linear (arithmetic) sequence. It goes up in 3s, so the common difference is $d = 3$.

Write down the multiples of the common difference

Because the common difference is 3, the nth term must start with 3n. Working out 3n for $n = 1$, 2, 3, 4 gives 3, 6, 9, 12.

Compare the multiples with the sequence

Every term is 5 more than the matching multiple of 3, so the constant is 5.

Write down the nth term

Putting the 3n and the constant together gives the rule 3n + 5.

Check the rule against the given terms

Substituting $n = 1$ gives 8 and $n = 4$ gives 17, which are the first and last terms given. The rule is correct - always test $n = 1$, because starting the count at the wrong place is the most common mistake.

Work out the 15th term

The 15th term is 50.

Work out the 40th term

The 40th term is 125.

Combine the two terms

Subtracting gives 75.

Check the answer a second way

The terms are 25 places apart, and each step adds 3, so the difference must be 25 lots of $3 = 75$. The two methods agree.

State the final answer

The required value is 75.

Find the common difference of sequence A

A has common difference 6.

Deduce the nth term of sequence A

6n gives 6,\ 12,\ 18,\ 24, and each term of A is 2 less, so $A = 6n - 2$.

Check the rule for A

It reproduces 4 and 22, the first and fourth terms of A.

Find the common difference of sequence B

B has common difference 2.

Deduce the nth term of sequence B

2n gives 2,\ 4,\ 6,\ 8, and each term of B is 30 more, so $B = 2n + 30$.

Check the rule for B

It reproduces 32 and 38, the first and fourth terms of B.

Set up the inequality

The question asks when a term of A first beats the term of B in the SAME position, so compare the two rules at the same n.

Collect the n terms on one side

Subtracting 2n and -2 from both sides gives $4n > 32$.

Simplify

The n terms combine into 4n.

Divide to find the boundary

So n must be greater than 8.

Round up to the first whole position

A position must be a whole number, and it must be strictly greater than 8, so the first one that works is $n = 9$.

Check position 8

At position 8, A gives 46 and B gives 46, so A is not yet ahead (46 is not greater than 46).

Check position 9

At position 9, A gives 52 and B gives 48, and 52 is greater than 48. This is the first position where A overtakes B.

Sense check the gap between the sequences

The gap A - B is 4n - 32, which increases by 4 each position - so once A is ahead it stays ahead.

State the answer

The first position at which the term of A is greater than the term of B is $n = 9$.