Arithmetic Sequences

Expert reviewed 21 July 2024 5 minute read


HSC Maths Advanced Syllabus

  • recognise and use the recursive definition of an arithmetic sequence: Tn=Tn1+dT_n = T_{n-1} + d, T1=aT_1 = a
  • establish and use the formula for the nthn^{th} term (where nn is a positive integer) of an arithmetic sequence: Tn=a+(n1)dT_n = a + (n - 1)d, where aa is the first term and dd is the common difference, and recognise its linear nature

Note: ‘Arithmetic Sequence’ is a term synonymous with ‘Arithmetic Progression’ (AP is the shorthand).

What is an Arithmetic Sequence?

An arithmetic sequence or AP for short, is a sequence in which the difference between consecutive terms is constant. This constant difference is known as the "common difference" represented by the letter d

The sequence progresses by adding (or subtracting, if d is negative) the common difference d to each term to get the next term.

Finding the nth term of an AP

The formula to find the nth term of an arithmetic sequence is given by:

Tn=a+(n1)dT_n=a+(n-1)d

Where aa is the first term in the sequence (T1)(T_1) and dd is the common difference.

Practice Question 1

Find a formula for the nth term of the arithmetic sequence: 3,6,9,12,...3,6,9,12,...

From the information given, we know that a = 3 and d = 3. Thus,

Tn=3+(n1)3=3+3n3=3nT_n = 3+(n-1)3\\=3+3n-3\\=3n

Practice Question 2

How many terms does the arithmetic series 3,6,9,...,453,6,9,...,45 contain?

From the previous question, we know that the formula for the nth term of this sequence is:

Tn=3nT_n = 3n

Thus, to find n we must equate TnT_n with the last term of the sequence.

Tn=453n=45n=15T_n = 45\\3n=45\\n=15

Thus, there are 15 terms in the sequence

Can an AP be formed using 3 numbers?

If the terms a, b and c are used in a sequence, as shown below:

a,b,ca,b,c

We can use the following formulas to determine if the sequence is an AP, and if it is they can be used to find missing terms.

ba=cbb=12(a+c)b-a = c-b\qquad \quad b=\frac{1}{2}(a+c)

Practice Question 3

Find the value of b, if the terms 15,b,2515,b,25 form an AP

b=12(15+25)=12(40)=20b = \frac{1}{2}(15+25)\\=\frac{1}{2}(40)\\=20

Return to Module 1: Sequences and Series