Arithmetic Progressions | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

A list or a sum?

A is an ordered list of numbers. Each number is a term. A adds those terms.

The small index in $u_n$ gives the position: $u_3$ is the third term, not $u\times3$ . $S_n$ is the sum of the first $n$ terms. Here $n=1,2,3,\ldots$ ; the dots mean the pattern continues.

Position	Term
1	5
2	8
3	11
4	14

S₄ = 5 + 8 + 11 + 14

= 38

u₃ = 11 is one term. S₄ = 38 adds four terms.

Add the same amount each time

In an (AP), each term changes by the same amount. Subtract a term from the next to find the common difference $d$ .

For $4,9,14,19,\ldots$ , $d=5$ . For $10,7,4,1,\ldots$ , $d=-3$ . A constant sequence has $d=0$ . The differences in $2,4,8,16,\ldots$ change, so it is not an AP.

Let $a$ be the first term. To reach term $n$ , make $n-1$ additions:

Term 1a
↓ add d
Term 2a + d
↓ add d
Term 3a + 2d

Three terms contain two changes.

u_n=a+(n-1)d

Check with $n=1$ : no change has happened, so $u_1=a$ . The exam formula booklet (MF19) supplies the AP term and sum formulae; you need to identify their inputs.

Find missing terms or positions

Worked example

An AP has $u_4=7$ and $u_{10}=25$ . Find $a$ and $d$ .

Translate both positions into equations:

a+3d=7,\qquad a+9d=25

Subtract the first equation from the second to remove $a$ . Then $6d=18$ , so $d=3$ and $a=7-3(3)=-2$ . Substitute both back to check.

To locate a value, solve the term equation for $n$ . In $4,9,14,\ldots$ , $4+5(n-1)=49$ gives $n=10$ . Replacing 49 by 50 gives $n=10.2$ , so 50 is not a term: positions must be positive integers.

If you are given $u_n=5-6n$ , substitute $n=1$ first. The first term is −1, not 5; the difference is −6.

Now you try

Your turn9709/12 Feb/Mar 2023 Q4(a) [2 marks]

A tree trunk has circumference 5.00 m, then 5.02 m one year later. Its yearly measurements form an AP. Find the circumference 20 years after the first measurement.

Show worked answer

The first measurement is at year zero. There are 20 changes, so use term 21, not term 20.

u_{21}=5.00+20(0.02)=5.40

Answer

5.40\text{ m}

Your turnIndependent practice

An AP has $u_3=8$ and $u_8=28$ . Find $u_n$ . Is 51 a term?

Show worked answer

The five changes total 20, so $d=4$ . Then $a+2(4)=8$ gives $a=0$ , and $u_n=4n-4$ .

$4n-4=51$ gives $n=13.75$ , not an integer. So 51 is not a term. Check: $u_{13}=48$ and $u_{14}=52$ .

A list or a sum?

A is an ordered list of numbers. Each number is a term. A adds those terms.

The small index in $u_n$ gives the position: $u_3$ is the third term, not $u\times3$ . $S_n$ is the sum of the first $n$ terms. Here $n=1,2,3,\ldots$ ; the dots mean the pattern continues.

Position	Term
1	5
2	8
3	11
4	14

S₄ = 5 + 8 + 11 + 14

= 38

u₃ = 11 is one term. S₄ = 38 adds four terms.

Add the same amount each time

In an (AP), each term changes by the same amount. Subtract a term from the next to find the common difference $d$ .

For $4,9,14,19,\ldots$ , $d=5$ . For $10,7,4,1,\ldots$ , $d=-3$ . A constant sequence has $d=0$ . The differences in $2,4,8,16,\ldots$ change, so it is not an AP.

Let $a$ be the first term. To reach term $n$ , make $n-1$ additions:

Term 1a
↓ add d
Term 2a + d
↓ add d
Term 3a + 2d

Three terms contain two changes.

u_n=a+(n-1)d

Check with $n=1$ : no change has happened, so $u_1=a$ . The exam formula booklet (MF19) supplies the AP term and sum formulae; you need to identify their inputs.

Find missing terms or positions

Worked example

An AP has $u_4=7$ and $u_{10}=25$ . Find $a$ and $d$ .

Translate both positions into equations:

a+3d=7,\qquad a+9d=25

Subtract the first equation from the second to remove $a$ . Then $6d=18$ , so $d=3$ and $a=7-3(3)=-2$ . Substitute both back to check.

To locate a value, solve the term equation for $n$ . In $4,9,14,\ldots$ , $4+5(n-1)=49$ gives $n=10$ . Replacing 49 by 50 gives $n=10.2$ , so 50 is not a term: positions must be positive integers.

If you are given $u_n=5-6n$ , substitute $n=1$ first. The first term is −1, not 5; the difference is −6.

Now you try

Your turn9709/12 Feb/Mar 2023 Q4(a) [2 marks]

A tree trunk has circumference 5.00 m, then 5.02 m one year later. Its yearly measurements form an AP. Find the circumference 20 years after the first measurement.

Show worked answer

The first measurement is at year zero. There are 20 changes, so use term 21, not term 20.

u_{21}=5.00+20(0.02)=5.40

Answer

5.40\text{ m}

Your turnIndependent practice

An AP has $u_3=8$ and $u_8=28$ . Find $u_n$ . Is 51 a term?

Show worked answer

The five changes total 20, so $d=4$ . Then $a+2(4)=8$ gives $a=0$ , and $u_n=4n-4$ .

$4n-4=51$ gives $n=13.75$ , not an integer. So 51 is not a term. Check: $u_{13}=48$ and $u_{14}=52$ .