Mathematics 9709 · Pure Mathematics 1

Arithmetic progressions

An arithmetic progression (AP) adds the same amount each step, like $5,\,8,\,11,\,14,\dots$ (up by $3$ ).
That fixed step is the common difference $d$ . The start is the first term $a$ .
Everything comes from two formulas. The skill is picking the right one.

Any term: $u_n=a+(n-1)d$ (you add $d$ one time fewer than $n$ ). Any sum: $S_n=\tfrac{n}{2}\big(2a+(n-1)d\big)$ . If you are given two terms, make two equations. Then subtract them to get $d$ .

The term and the sum

To get to the $n$ th term, you start at $a$ and add $d$ a total of $(n-1)$ times.
That is one time fewer than $n$ , because the first term is already there before you add anything.

$u_n=a+(n-1)d$

The sum of the first $n$ terms has its own formula (use the $a+l$ form when you know the last term $l$ ):

$S_n=\tfrac{n}{2}\big(a+l\big)=\tfrac{n}{2}\big(2a+(n-1)d\big)$

Both are in your MF19 formula list. So you only have to choose between them. You do not have to learn them.
They are the same formula. The last term is $l=a+(n-1)d$ . That is why $a+l$ becomes $2a+(n-1)d$ .

Where the sum formula comes from (Gauss's method)

Write the sum forwards. Then write it again backwards underneath. Now add the two lines:

$S_n=\;\,a\;+\;(a+d)\;+\;\dots\;+\;l$

$S_n=\;\,l\;+\;(l-d)\;+\;\dots\;+\;a$

Each column now adds to the same total, the first plus the last, $a+l$ . The $+d$ in one row cancels the $-d$ in the other. There are $n$ columns, so $2S_n=n(a+l)$ . This gives $S_n=\tfrac{n}{2}(a+l)$ . The story is that the young Gauss used this to add $1$ to $100$ in seconds.

Worked examples

You will do this again and again. Turn each fact into an equation in $a$ and $d$ . Then subtract to remove $a$ and leave the step $d$ on its own.

Worked example

Two given terms give $a$ and $d$

The 4th term of an AP is $7$ and the 10th is $16$ .

Turn each term into an equation with $u_n=a+(n-1)d$ : $a+3d=7$ and $a+9d=16$ .
Subtract the first from the second to remove $a$ : $6d=9\Rightarrow d=\tfrac32$ .
Put $d$ back into either equation: $a+\tfrac92=7$ .

Answer

a=\tfrac52,\quad d=\tfrac32

Worked example

Real exam: sum the first 20 terms, then a condition on $2k$ and $k$

The first term of an AP is $-20$ and the common difference is $5$ .

(a) Find the sum of the first 20 terms. Here $a=-20$ , $d=5$ , $n=20$ . Put them into $S_n=\tfrac{n}{2}(2a+(n-1)d)$ :

$S_{20}=\tfrac{20}{2}\big(2(-20)+19\times5\big)=10\,(-40+95)=10\times55$

Answer

S_{20}=550

(b) The sum of the first $2k$ terms is $10$ times the sum of the first $k$ terms. Find $k$ . Write each sum with the same formula. Use $2k$ terms for one, and $k$ terms for the other:

$S_{2k}=\tfrac{2k}{2}\big(-40+(2k-1)5\big)=k(10k-45)$

$S_{k}=\tfrac{k}{2}\big(-40+(k-1)5\big)=\tfrac{k}{2}(5k-45)$

Set $S_{2k}=10\,S_k$ . Then divide both sides by $k$ (it counts terms, so it is not zero):

$10k-45=5(5k-45)\Rightarrow 10k-45=25k-225$

This one stays linear: $180=15k$ .

Answer

k=12

(9709/12 Nov 2024 Q2) Dividing by $k$ is the tidiest way here. You can multiply it all out instead, but that makes sign slips more likely.

Worked example

A known last term: count the terms, then sum with $\tfrac{n}{2}(a+l)$

An AP is $-3,\,1,\,5,\,9,\dots,\,237$ . Find how many terms it has. Then find the sum of them all. Here $a=-3$ , $d=4$ , and the last term $l=237$ is given.

The last term is the $n$ th term. So set $u_n=a+(n-1)d$ equal to $237$ , then solve for $n$ :

$237=-3+(n-1)4 \;\Rightarrow\; (n-1)4=240 \;\Rightarrow\; n=61$

You now know the first term, the last term and how many terms there are. So the $\tfrac{n}{2}(a+l)$ form is the quick one to use:

$S_{61}=\tfrac{61}{2}\big(a+l\big)=\tfrac{61}{2}(-3+237)=\tfrac{61}{2}\times 234$

Answer

n=61,\quad S_{61}=7137

Worked example

Work backwards from the $n$ th-term rule

The $n$ th term of an AP is $\tfrac12(3n-15)$ , and the sum of the first $n$ terms is $84$ . Find $n$ . (9709/12 Jun 2020 Q4)

First get $a$ and $d$ from the rule. Put $n=1$ for the first term. Put $n=2$ to find the step:

$a=\tfrac12(3-15)=-6,\qquad u_2=\tfrac12(6-15)=-\tfrac92$

So $d=u_2-a=-\tfrac92-(-6)=\tfrac32$ .

Now put those into $S_n=84$ :

$\tfrac{n}{2}\big(2(-6)+(n-1)\tfrac32\big)=84$

Here the question becomes a quadratic in $n$ . Multiply out and clear the fractions (times $4$ ):

$3n^2-27n-336=0 \;\Rightarrow\; n^2-9n-112=0$

Solve it the calculator-first way. Type the quadratic into your calculator's equation solver. It gives $n=16$ and $n=-7$ . (If you like factors, those roots come from $(n-16)(n+7)=0$ . The Quadratics chapter has the full method.) A count of terms cannot be negative, so reject $n=-7$ :

Answer

n=16

Three consecutive AP terms: the middle is the average

When a question gives three terms in a row $a,\,b,\,c$ , you do not need $a$ and $d$ at all. The gaps are equal, so $b-a=c-b$ . This means the middle term is the average of the other two:

$2b=a+c$

For example, say $k,\,6k,\,k+6$ are three AP terms in a row. Then $2(6k)=k+(k+6)$ , which gives $12k=2k+6\Rightarrow k=\tfrac35$ . One line, no formula. (9709/12 Jun 2022 Q4)

Where the marks go

Common mistake

The

n

th term uses

(n-1)d

, not

nd

. So the 4th term is

a+3d

, never

a+4d

. This off-by-one is the most common AP slip.

Common mistake

A sum question often becomes a quadratic in $n$ . Solve it. Then reject any root that is negative or not a whole number. Here

n

counts terms, so it must be a positive whole number. (9709/12 Jun 2020 Q4)

Now you try

The first, second and third terms of an AP are $k$ , $6k$ and $k+6$ respectively. Find $k$ , then the sum of the first $30$ terms. (9709/12 Jun 2022 Q4)

Your turn— tap to reveal the worked answer (9709/12 Jun 2022 Q4)

Find $k$ . These are three AP terms in a row. So the middle is the average (both gaps equal $d$ ). Use $2b=a+c$ :

$2(6k)=k+(k+6)\Rightarrow 12k=2k+6\Rightarrow k=\tfrac35$

Sum the first 30. Now the first term is $a=k=\tfrac35$ and the common difference is $d=6k-k=5k=3$ :

$S_{30}=\tfrac{30}{2}\big(2\cdot\tfrac35+29\cdot3\big)=15\big(\tfrac65+87\big)$

Answer

k=\tfrac35,\quad S_{30}=1323

PreviousThe binomial expansion NextGeometric progressions

Arithmetic progressions

An arithmetic progression (AP) adds the same amount each step, like $5,\,8,\,11,\,14,\dots$ (up by $3$ ).
That fixed step is the common difference $d$ . The start is the first term $a$ .
Everything comes from two formulas. The skill is picking the right one.

Any term: $u_n=a+(n-1)d$ (you add $d$ one time fewer than $n$ ). Any sum: $S_n=\tfrac{n}{2}\big(2a+(n-1)d\big)$ . If you are given two terms, make two equations. Then subtract them to get $d$ .

The term and the sum

To get to the $n$ th term, you start at $a$ and add $d$ a total of $(n-1)$ times.
That is one time fewer than $n$ , because the first term is already there before you add anything.

$u_n=a+(n-1)d$

The sum of the first $n$ terms has its own formula (use the $a+l$ form when you know the last term $l$ ):

$S_n=\tfrac{n}{2}\big(a+l\big)=\tfrac{n}{2}\big(2a+(n-1)d\big)$

Both are in your MF19 formula list. So you only have to choose between them. You do not have to learn them.
They are the same formula. The last term is $l=a+(n-1)d$ . That is why $a+l$ becomes $2a+(n-1)d$ .

Where the sum formula comes from (Gauss's method)

Write the sum forwards. Then write it again backwards underneath. Now add the two lines:

$S_n=\;\,a\;+\;(a+d)\;+\;\dots\;+\;l$

$S_n=\;\,l\;+\;(l-d)\;+\;\dots\;+\;a$

Worked examples

You will do this again and again. Turn each fact into an equation in $a$ and $d$ . Then subtract to remove $a$ and leave the step $d$ on its own.

Worked example

Two given terms give $a$ and $d$

The 4th term of an AP is $7$ and the 10th is $16$ .

Turn each term into an equation with $u_n=a+(n-1)d$ : $a+3d=7$ and $a+9d=16$ .
Subtract the first from the second to remove $a$ : $6d=9\Rightarrow d=\tfrac32$ .
Put $d$ back into either equation: $a+\tfrac92=7$ .

Answer

a=\tfrac52,\quad d=\tfrac32

Worked example

Real exam: sum the first 20 terms, then a condition on $2k$ and $k$

The first term of an AP is $-20$ and the common difference is $5$ .

(a) Find the sum of the first 20 terms. Here $a=-20$ , $d=5$ , $n=20$ . Put them into $S_n=\tfrac{n}{2}(2a+(n-1)d)$ :

$S_{20}=\tfrac{20}{2}\big(2(-20)+19\times5\big)=10\,(-40+95)=10\times55$

Answer

S_{20}=550

(b) The sum of the first $2k$ terms is $10$ times the sum of the first $k$ terms. Find $k$ . Write each sum with the same formula. Use $2k$ terms for one, and $k$ terms for the other:

$S_{2k}=\tfrac{2k}{2}\big(-40+(2k-1)5\big)=k(10k-45)$

$S_{k}=\tfrac{k}{2}\big(-40+(k-1)5\big)=\tfrac{k}{2}(5k-45)$

Set $S_{2k}=10\,S_k$ . Then divide both sides by $k$ (it counts terms, so it is not zero):

$10k-45=5(5k-45)\Rightarrow 10k-45=25k-225$

This one stays linear: $180=15k$ .

Answer

k=12

(9709/12 Nov 2024 Q2) Dividing by $k$ is the tidiest way here. You can multiply it all out instead, but that makes sign slips more likely.

Worked example

A known last term: count the terms, then sum with $\tfrac{n}{2}(a+l)$

An AP is $-3,\,1,\,5,\,9,\dots,\,237$ . Find how many terms it has. Then find the sum of them all. Here $a=-3$ , $d=4$ , and the last term $l=237$ is given.

The last term is the $n$ th term. So set $u_n=a+(n-1)d$ equal to $237$ , then solve for $n$ :

$237=-3+(n-1)4 \;\Rightarrow\; (n-1)4=240 \;\Rightarrow\; n=61$

You now know the first term, the last term and how many terms there are. So the $\tfrac{n}{2}(a+l)$ form is the quick one to use:

$S_{61}=\tfrac{61}{2}\big(a+l\big)=\tfrac{61}{2}(-3+237)=\tfrac{61}{2}\times 234$

Answer

n=61,\quad S_{61}=7137

Worked example

Work backwards from the $n$ th-term rule

The $n$ th term of an AP is $\tfrac12(3n-15)$ , and the sum of the first $n$ terms is $84$ . Find $n$ . (9709/12 Jun 2020 Q4)

First get $a$ and $d$ from the rule. Put $n=1$ for the first term. Put $n=2$ to find the step:

$a=\tfrac12(3-15)=-6,\qquad u_2=\tfrac12(6-15)=-\tfrac92$

So $d=u_2-a=-\tfrac92-(-6)=\tfrac32$ .

Now put those into $S_n=84$ :

$\tfrac{n}{2}\big(2(-6)+(n-1)\tfrac32\big)=84$

Here the question becomes a quadratic in $n$ . Multiply out and clear the fractions (times $4$ ):

$3n^2-27n-336=0 \;\Rightarrow\; n^2-9n-112=0$

Answer

n=16

Three consecutive AP terms: the middle is the average

When a question gives three terms in a row $a,\,b,\,c$ , you do not need $a$ and $d$ at all. The gaps are equal, so $b-a=c-b$ . This means the middle term is the average of the other two:

$2b=a+c$

For example, say $k,\,6k,\,k+6$ are three AP terms in a row. Then $2(6k)=k+(k+6)$ , which gives $12k=2k+6\Rightarrow k=\tfrac35$ . One line, no formula. (9709/12 Jun 2022 Q4)

Where the marks go

Common mistake

The

n

th term uses

(n-1)d

, not

nd

. So the 4th term is

a+3d

, never

a+4d

. This off-by-one is the most common AP slip.

Common mistake

A sum question often becomes a quadratic in $n$ . Solve it. Then reject any root that is negative or not a whole number. Here

n

counts terms, so it must be a positive whole number. (9709/12 Jun 2020 Q4)

Now you try

The first, second and third terms of an AP are $k$ , $6k$ and $k+6$ respectively. Find $k$ , then the sum of the first $30$ terms. (9709/12 Jun 2022 Q4)

Your turn— tap to reveal the worked answer (9709/12 Jun 2022 Q4)

Find $k$ . These are three AP terms in a row. So the middle is the average (both gaps equal $d$ ). Use $2b=a+c$ :

$2(6k)=k+(k+6)\Rightarrow 12k=2k+6\Rightarrow k=\tfrac35$

Sum the first 30. Now the first term is $a=k=\tfrac35$ and the common difference is $d=6k-k=5k=3$ :

$S_{30}=\tfrac{30}{2}\big(2\cdot\tfrac35+29\cdot3\big)=15\big(\tfrac65+87\big)$

Answer

k=\tfrac35,\quad S_{30}=1323