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Study Guide For Grade

12

Mathematics

Term 4
Paper 1

Patterns, Sequences and Series

Number patterns

A number pattern is an ordered list of numbers that follows a rule. Each number is called a term. The position of a term is represented by n, while the term in that position is written as T_n.

To identify a pattern:

  • Calculate the first differences between consecutive terms.
  • If these are constant, the pattern is arithmetic.
  • If each term is multiplied by the same number, the pattern is geometric.
  • If the second differences are constant, the pattern is quadratic.

Example

Consider the pattern:

3; 7; 11; 15;…3;\ 7;\ 11;\ 15;\ldots

The first differences are:

7−3=47-3=4
11−7=411-7=4
15−11=415-11=4

Therefore, the pattern is arithmetic with a common difference of 4.

Remember: A sequence is a list of terms, while a series is the sum of the terms.

Arithmetic sequences

In an arithmetic sequence, the difference between consecutive terms is constant. This value is called the common difference, d.

d=Tn+1−Tnd=T_{n+1}-T_n

The general term is:

Tn=a+(n−1)dT_n=a+(n-1)d

Here, a is the first term, d is the common difference and n is the term number.

Example

Determine the 20th term of:

3; 7; 11;…3;\ 7;\ 11;\ldots
a=3a=3
d=7−3=4d=7-3=4
Tn=a+(n−1)dT_n=a+(n-1)d
T20=3+(20−1)(4)T_{20}=3+(20-1)(4)
T20=3+76T_{20}=3+76
T20=79T_{20}=79

An important property of an arithmetic sequence is that terms equally far from a central term have a constant difference determined by their positions. For example, if the common difference is 4:

T14−T6=(14−6)(4)=32T_{14}-T_6=(14-6)(4)=32

This is a useful shortcut in examination questions.

Geometric sequences

In a geometric sequence, each term is obtained by multiplying the previous term by a constant called the common ratio, r.

r=Tn+1Tnr=\frac{T_{n+1}}{T_n}

The general term is:

Tn=arn−1T_n=ar^{n-1}

Example

Determine the 25th term of:

−8; 4; −2; 1;…-8;\ 4;\ -2;\ 1;\ldots
a=−8a=-8
r=4−8=−12r=\frac{4}{-8}=-\frac{1}{2}
Tn=arn−1T_n=ar^{n-1}
T25=−8(−12)24T_{25}=-8\left(-\frac{1}{2}\right)^{24}
T25=−8(12)24T_{25}=-8\left(\frac{1}{2}\right)^{24}
T25=−1221T_{25}=-\frac{1}{2^{21}}

This example shows why the sign of the exponent must be handled carefully when the common ratio is negative.

Quadratic number patterns

A quadratic number pattern has constant second differences. Its general term has the form:

Tn=an2+bn+cT_n=an^2+bn+c

The constant second difference equals 2a.

Example

Determine the general term of:

3; 8; 15; 24;…3;\ 8;\ 15;\ 24;\ldots

First differences:

5; 7; 95;\ 7;\ 9

Second differences:

2; 22;\ 2

Therefore:

2a=22a=2
a=1a=1

Use the first two terms to determine b and c.

T1=1+b+c=3T_1=1+b+c=3
b+c=2b+c=2
T2=4+2b+c=8T_2=4+2b+c=8
2b+c=42b+c=4

Subtract the equations:

b=2b=2
c=0c=0

Therefore:

Tn=n2+2nT_n=n^2+2n

General term of a sequence

The general term gives the value of any term directly without listing all the preceding terms.

For an arithmetic sequence:

Tn=a+(n−1)dT_n=a+(n-1)d

For a geometric sequence:

Tn=arn−1T_n=ar^{n-1}

For a quadratic sequence:

Tn=an2+bn+cT_n=an^2+bn+c

Example

Determine which term of the sequence below is 467.

3; 7; 11;…3;\ 7;\ 11;\ldots
a=3a=3
d=4d=4
Tn=a+(n−1)dT_n=a+(n-1)d
467=3+(n−1)(4)467=3+(n-1)(4)
467=3+4n−4467=3+4n-4
467=4n−1467=4n-1
468=4n468=4n
n=117n=117

Therefore, 467 is the 117th term.

Arithmetic series

An arithmetic series is the sum of the terms of an arithmetic sequence.

The sum of the first n terms is:

Sn=n2[2a+(n−1)d]S_n=\frac{n}{2}\left[2a+(n-1)d\right]

Alternatively, if the last term is known:

Sn=n2(a+Tn)S_n=\frac{n}{2}(a+T_n)

Example

Calculate the sum of the first 20 terms of:

3+7+11+…3+7+11+\ldots
a=3a=3
d=4d=4
n=20n=20
Sn=n2[2a+(n−1)d]S_n=\frac{n}{2}\left[2a+(n-1)d\right]
S20=202[2(3)+(20−1)(4)]S_{20}=\frac{20}{2}\left[2(3)+(20-1)(4)\right]
S20=10(6+76)S_{20}=10(6+76)
S20=820S_{20}=820

Geometric series

A geometric series is formed when the terms of a geometric sequence are added.

a+ar+ar2+ar3+…a+ar+ar^2+ar^3+\ldots

The formula used depends on whether the series is finite or infinite.

Always determine a and r before selecting a formula. If the signs alternate, the common ratio is negative.

Finite geometric series

A finite geometric series has a fixed number of terms. Its sum is:

Sn=a(rn−1)r−1S_n=\frac{a(r^n-1)}{r-1}

Alternatively:

Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}

These formulas apply when:

r≠1r\ne 1

Example

Calculate the sum of the first five terms of:

2+6+18+…2+6+18+\ldots
a=2a=2
r=62=3r=\frac{6}{2}=3
Sn=a(rn−1)r−1S_n=\frac{a(r^n-1)}{r-1}
S5=2(35−1)3−1S_5=\frac{2(3^5-1)}{3-1}
S5=2(243−1)2S_5=\frac{2(243-1)}{2}
S5=242S_5=242

Infinite geometric series

An infinite geometric series continues without ending:

a+ar+ar2+ar3+…a+ar+ar^2+ar^3+\ldots

A sum to infinity exists only if:

−1<r<1-1<r<1

Equivalently:

|r|<1\lvert r\rvert<1

If the absolute value of r is greater than or equal to 1, the terms do not approach zero and the series has no sum to infinity.

Sum to infinity

The sum to infinity of a convergent geometric series is:

S∞=a1−rS_{\infty}=\frac{a}{1-r}

Example

Calculate the sum to infinity of:

−8+4−2+1−…-8+4-2+1-\ldots
a=−8a=-8
r=4−8=−12r=\frac{4}{-8}=-\frac{1}{2}
|−12|<1\left|-\frac{1}{2}\right|<1

Therefore, the series converges.

S∞=a1−rS_{\infty}=\frac{a}{1-r}
S∞=−81−(−12)S_{\infty}=\frac{-8}{1-\left(-\frac{1}{2}\right)}
S∞=−832S_{\infty}=\frac{-8}{\frac{3}{2}}
S∞=−163S_{\infty}=-\frac{16}{3}

This substitution and answer agree with the expected examination method.

Sigma notation

Sigma notation is a compact way of writing a series. The Greek capital letter sigma represents summation.

∑k=1nTk\sum_{k=1}^{n}T_k

The lower value gives the starting value of the index, and the upper value gives the final value.

Example

Expand:

∑k=14(2k+1)\sum_{k=1}^{4}(2k+1)
=[2(1)+1]+[2(2)+1]+[2(3)+1]+[2(4)+1]=[2(1)+1]+[2(2)+1]+[2(3)+1]+[2(4)+1]
=3+5+7+9=3+5+7+9
=24=24

Useful sigma results include:

∑k=1nc=cn\sum_{k=1}^{n}c=cn
∑k=1nk=n(n+1)2\sum_{k=1}^{n}k=\frac{n(n+1)}{2}
∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}

Determining unknown terms and constants

Unknown values are found by applying the defining property of the sequence or series.

Example: Unknown in an arithmetic sequence

The following are consecutive terms of an arithmetic sequence:

x−3; 2x+1; 5x−7x-3;\ 2x+1;\ 5x-7

Consecutive differences must be equal.

(2x+1)−(x−3)=(5x−7)−(2x+1)(2x+1)-(x-3)=(5x-7)-(2x+1)
x+4=3x−8x+4=3x-8
12=2x12=2x
x=6x=6

The terms are therefore:

3; 13; 233;\ 13;\ 23

Example: Unknown in a geometric sequence

Suppose 2, k and 18 are consecutive terms. Equal ratios give:

k2=18k\frac{k}{2}=\frac{18}{k}
k2=36k^2=36
k=±6k=\pm 6

Both values must be considered unless the question states that the terms are positive.

Exam Tip

Write down the correct formula before substituting. Show the number of terms, common difference or common ratio, substitution, simplification and final answer. When solving for an index, reject negative or non-integer values because a term number must be a positive integer. Examination marking commonly awards separate marks for these stages and for selecting the valid solution.