What comes after 16 in the sequence 1, 4, 16, 64, 256?

64. This is a geometric sequence: each term is the one before multiplied by 4, since 1 × 4 = 4, 4 × 4 = 16, 16 × 4 = 64 and 64 × 4 = 256. The ratio is 4, not a constant difference, so the term after 16 is 16 × 4 = 64. Treating it as a difference sequence fails immediately, because the gaps 3, 12, 48, 192 are not equal.

How do you continue a Fibonacci sequence like 5, −9?

Each term after the first two is the sum of the two terms before it. Starting 5, −9, the next term is 5 + (−9) = −4, then (−9) + (−4) = −13, so the sequence is 5, −9, −4, −13. You add the two previous terms each time, keeping the signs; you do not add a constant difference.

How do you tell if a sequence is arithmetic, geometric or Fibonacci?

Check the rule before continuing. If each term is the previous one plus a fixed amount, it is arithmetic (constant difference). If each term is the previous one times a fixed amount, it is geometric (constant ratio), as in 1, 4, 16, 64 with ratio 4. If each term is the sum of the two before it, it is Fibonacci-style. Test the first few terms against each rule rather than assuming a constant difference.

GCSE Maths Foundation · AQA · Sequences

Geometric and Fibonacci sequences: check the rule before you continue

Most sequences students meet first go up by a constant difference, so that habit gets applied to every sequence. But $1,\ 4,\ 16,\ 64,\ 256$ is geometric — each term is the one before multiplied by 4 — and a Fibonacci sequence like $5,\ -9,\ \dots$ adds the two previous terms. Reading either as “add the same amount” gives the wrong terms.

The thirty-second fix: check the rule before continuing — a geometric sequence multiplies by a fixed ratio, and a Fibonacci sequence adds the two terms before. So the term after 16 in $1,\ 4,\ 16,\ 64,\ 256$ is $16 \times 4 = 64$ , and $5,\ -9$ continues $5 + (-9) = -4$ , then $(-9) + (-4) = -13$ .

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How to spot it in your own work

You assumed a constant difference, so for $1,\ 4,\ 16,\ 64,\ 256$ you looked for “add the same amount” when the rule is × 4.
You did not test the gaps: $3,\ 12,\ 48,\ 192$ are not equal, so the sequence is not arithmetic.
You continued a Fibonacci sequence by adding a fixed number instead of summing the two previous terms, missing $5 + (-9) = -4$ .
You dropped or mishandled the sign while summing terms, e.g. treating $(-9) + (-4)$ as $-5$ instead of $-13$ .

An exam question that triggers it

Here is a canonical AQA Foundation trigger (non-calculator paper, find the missing term):

Here is a sequence.

$1,\quad 4,\quad 16,\quad \square,\quad 256$

Work out the missing term.

The misconception is to hunt for a constant difference. The gaps $3,\ 12,\ \dots$ are not equal, so there is no fixed difference — the rule is multiplicative.

Each term is the one before times 4: $1 \times 4 = 4$ , $4 \times 4 = 16$ , so the missing term is $16 \times 4 = 64$ (and $64 \times 4 = 256$ confirms it).

Why students fall for this

The first sequences anyone studies — $2,\ 5,\ 8,\ 11$ , the multiples, the linear ones — all go up by a fixed amount, so “find the difference and keep adding” becomes the default move. Applied to $1,\ 4,\ 16,\ 64,\ 256$ it fails at once: the gaps are $3,\ 12,\ 48,\ 192$ , which grow rather than stay equal. The pattern is multiplicative — each term is the previous one times the same ratio, here 4.

Fibonacci-style sequences break the habit a second way: there is no single number that is added at all. Each term is the sum of the two before it. Starting $5,\ -9$ , the next term is $5 + (-9) = -4$ , then $(-9) + (-4) = -13$ , giving $5,\ -9,\ -4,\ -13$ — and the signs have to be carried carefully, since adding two negatives makes a larger negative.

AQA Foundation papers exploit both directly: a geometric run with a blank to fill, and a Fibonacci-style sequence where you continue from two given terms. Assuming a constant difference produces a confident wrong answer in each case.

Worked example — a Fibonacci continuation. Continue $5,\ -9$ for two more terms.

Each term is the sum of the two before it, keeping the signs:

5 + (-9) = -4 \qquad (-9) + (-4) = -13

So the sequence runs $5,\ -9,\ -4,\ -13$ . Adding a fixed difference would miss this entirely — there is no constant being added.

The fix: Check the rule first: geometric multiplies by a fixed ratio, Fibonacci adds the two previous terms

Test for a constant difference first. Find the gaps. For $1,\ 4,\ 16,\ 64,\ 256$ they are $3,\ 12,\ 48,\ 192$ — not equal, so it is not arithmetic.

Then test for a constant ratio. Divide each term by the one before: $4 \div 1 = 4$ , $16 \div 4 = 4$ — a fixed ratio of 4, so it is geometric. The missing term is $16 \times 4 = 64$ .

If neither, try adding the two previous terms. A Fibonacci-style sequence has each term equal to the sum of the two before: $5,\ -9$ gives $5 + (-9) = -4$ , then $(-9) + (-4) = -13$ .

Carry the signs. When the terms are negative, add them as directed numbers — $(-9) + (-4) = -13$ , a larger negative, not a smaller one.

Worked example

Fill the blank in $1,\ 4,\ 16,\ \square,\ 256$ , then continue $5,\ -9$ . The trap in both is to assume a constant difference; the fix is to identify the rule first.

Rule out a constant difference. The gaps in $1,\ 4,\ 16$ are $3$ then $12$ — not equal, so the sequence is not arithmetic.
Find the ratio. $4 \div 1 = 4$ and $16 \div 4 = 4$ , so each term is the previous one times 4 — geometric with ratio 4.
Fill the blank.
$16 \times 4 = 64 \qquad (64 \times 4 = 256\ \checkmark)$
The missing term is $64$ , confirmed by the 256 that follows.
Continue the Fibonacci sequence. Each term is the sum of the two before, carrying the signs:
$5 + (-9) = -4 \qquad (-9) + (-4) = -13$
so $5,\ -9$ continues $5,\ -9,\ -4,\ -13$ .

So before continuing any sequence, check whether the rule is add a fixed amount, multiply by a fixed ratio, or add the two previous terms — the missing term after 16 is $64$ , not whatever a constant difference would predict.

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Common questions

What comes after 16 in the sequence $1,\ 4,\ 16,\ 64,\ 256$ ?: 64. This is a geometric sequence: each term is the one before multiplied by 4, since $1 \times 4 = 4$ , $4 \times 4 = 16$ , $16 \times 4 = 64$ and $64 \times 4 = 256$ . The ratio is 4, not a constant difference, so the term after 16 is $16 \times 4 = 64$ . Treating it as a difference sequence fails immediately, because the gaps $3,\ 12,\ 48,\ 192$ are not equal.
How do you continue a Fibonacci sequence like $5,\ -9$ ?: Each term after the first two is the sum of the two terms before it. Starting $5,\ -9$ , the next term is $5 + (-9) = -4$ , then $(-9) + (-4) = -13$ , so the sequence is $5,\ -9,\ -4,\ -13$ . You add the two previous terms each time, keeping the signs; you do not add a constant difference.
How do you tell if a sequence is arithmetic, geometric or Fibonacci?: Check the rule before continuing. If each term is the previous one plus a fixed amount, it is arithmetic (constant difference). If each term is the previous one times a fixed amount, it is geometric (constant ratio), as in $1,\ 4,\ 16,\ 64$ with ratio 4. If each term is the sum of the two before it, it is Fibonacci-style. Test the first few terms against each rule rather than assuming a constant difference.

Related misconceptions

Finding the nth-term rule of a sequenceThe linear case where there is a constant difference: that difference becomes the coefficient of n, so 2, 5, 8, 11 is 3n − 1.
Term-to-term vs nth-term rulesWhich kind of rule the question wants: a term-to-term step like add 6, or an nth-term expression like 6n − 1 that reaches any term.

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