HA23–HA25 - Quadratic and geometric sequence rules | Eduqas GCSE Mathematics

The gaps in a sequence may change regularly even when the terms do not. How can that reveal a formula for a distant term? HA23–HA25 connects second differences to quadratic rules, checks every coefficient, and extends geometric patterns to exact surd ratios.

Second differences

For the sequence $3,8,15,24,35,\ldots$ , the first gaps are $5,7,9,11$ . Those gaps increase by two. This is the feature that distinguishes a quadratic pattern from a linear one.

Quadratic sequences

A quadratic sequence has a position rule $T_n=an^2+bn+c$ , with $a\ne0$ and positive integer positions $n=1,2,3,\ldots$ . A first difference is a term minus the preceding term. A second difference is the difference between consecutive first differences. For a quadratic sequence, the second difference is the constant $2a$ .

The link comes from the square term. The first differences of $n^2$ are $3,5,7,9,\ldots$ , with second difference two. Multiplying by $a$ multiplies that second difference by $a$ . A linear part $bn+c$ adds the same $b$ to each first difference, so contributes zero to second differences.

Algebraically,

T_{n+1}-T_n=a((n+1)^2-n^2)+b=2an+a+b.

Increasing $n$ by one increases this first difference by $2a$ . Thus $a$ is half the second difference, including when it is negative or fractional.

Worked example 1: positive second differences

Question. A quadratic sequence begins $3,8,15,24,35$ . Find its second difference, its coefficient of $n^2$ , and its next term.

Plan. Build both difference rows, then extend the first-difference pattern once.

Terms	$3$	$8$	$15$	$24$	$35$
First differences		$5$	$7$	$9$	$11$
Second differences			$2$	$2$	$2$

Since $2a=2$ , $a=1$ . The next first difference is $13$ , so the next term is $35+13=48$ .

Answer. Second difference two; coefficient one; next term $48$ .

Check. The candidate rule $T_n=n^2+2n$ produces the five supplied terms at positions one to five, and gives $T_6=36+12=48$ . This independently checks the extended term; the next part shows how to deduce the whole rule systematically.

Second differences can be negative, so a quadratic sequence need not keep increasing.

Worked example 2: negative second differences

Question. A quadratic sequence begins $10,14,12,4,-10$ . Find the coefficient of $n^2$ and the next term.

Plan. Subtract consecutive terms carefully, preserving the signs of the differences.

The first differences are $4,-2,-8,-14$ . Their differences are all $-6$ , so

2a=-6\quad\Rightarrow\quad a=-3.

The next first difference is $-14-6=-20$ , and the next term is $-10-20=-30$ .

Answer. Coefficient $-3$ ; next term $-30$ .

Check. The rule $-3n^2+13n$ gives $10,14,12,4,-10$ for the original five positions, and $-108+78=-30$ at position six. Both the change in direction and the negative next term are consistent with the rule.

When a question states that a sequence is quadratic, its constant second difference can be continued. A short list alone does not prove that every future term follows that pattern; a different specified rule could agree initially and then change. Also, a zero second difference belongs to a linear pattern rather than a genuine quadratic with $a\ne0$ .

Watch out

The second difference is $2a$ , not $a$ . Check the square sequence itself: second difference two corresponds to coefficient one.

Use both rows to extend a fresh quadratic pattern.

Quadratic nth terms

The second difference determines $a$ , but not $b$ or $c$ . Subtract the matching square contribution to reveal the remaining linear pattern.

Finding the whole quadratic rule

For a known quadratic sequence, first find $a$ from its second difference. Subtract $an^2$ from each original term. The remainders follow $bn+c$ : their common first difference is $b$ , and the remainder at $n=1$ is $b+c$ . Find $c$ , then combine all three terms and substitute into the original data to check.

Use positions starting at one unless the question specifies otherwise. Subtract the square contribution at each matching position, rather than subtracting the same number from every term.

Worked example 3: a positive linear remainder

Question. Find the nth term of the quadratic sequence $4,11,22,37,56,\ldots$ , then find its tenth term.

Plan. Find the square coefficient, subtract $an^2$ , and identify the remainder's linear rule.

First differences are $7,11,15,19$ ; second differences are four, so $a=2$ .

$n$	$1$	$2$	$3$	$4$	$5$
Original $T_n$	$4$	$11$	$22$	$37$	$56$
$2n^2$	$2$	$8$	$18$	$32$	$50$
Remainder	$2$	$3$	$4$	$5$	$6$

The remainder increases by one, so $b=1$ . At position one, $b+c=2$ , giving $c=1$ . The remainder is $n+1$ .

Answer. $T_n=2n^2+n+1$ ; $T_{10}=200+10+1=211$ .

Check. Independent substitution gives $T_1=4$ , $T_3=18+3+1=22$ , $T_5=50+5+1=56$ , matching the original positions. Continuing the differences predicts $56+23=79$ ; the formula independently gives $T_6=72+6+1=79$ .

The same second difference can accompany a very different linear remainder. That is why the square coefficient alone is insufficient.

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HA23–HA25 - Quadratic and geometric sequence rules

Second differences

Quadratic nth terms

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