Arithmetic Sequences For A Level Maths

Arithmetic Sequences For A Level Maths – Introduction

An arithmetic sequence in A Level Maths is a sequence of numbers in which the difference between consecutive terms remains constant. It can be represented by the formula:

a_n=a+(n-1) d

where:
$a_n$ = the nth term of the sequence
a = the first term of the sequence
n = the position of the term in the sequence
d = the common difference between consecutive terms

Properties of an arithmetic sequence:

Common Difference: The common difference (d) between consecutive terms remains constant throughout the sequence.
Constant Increment: Each term in the sequence increases or decreases by the same amount (d).
Linear Relationship: The relationship between the position of a term in the sequence and the value of the term is linear.

Finding the nth term of an arithmetic sequence:

To find the nth term $\left(a_n\right)$ of an arithmetic sequence, we use the formula mentioned earlier:

a_n=a+(n-1) d

where a is the first term and d is the common difference.

Finding the sum of an arithmetic sequence:
The sum of the first n terms of an arithmetic sequence can be calculated using the formula:

S_n=\frac{n}{2}[2 a+(n-1) d]

where $S_n$ represents the sum of the first n terms, a is the first term, n is the number of terms, and d is the common difference.

Example:
Let’s consider an arithmetic sequence with a first term (a) of 2 and a common difference (d) of 3. We want to find the value of the 7th term $\left(a_7\right)$ and the sum of the first 7 terms $\left(S_7\right)$ .

Using the formula for the nth term:

\begin{aligned} &\begin{aligned} & a_7=a+(7-1) d \\ & a_7=2+6(3) \end{aligned}\\ &\begin{aligned} & a_7=2+18 \\ & a_7=20 \end{aligned} \end{aligned}

Using the formula for the sum of the first n terms:

\begin{aligned} & S_7=(7 / 2)(2(2)+(7-1)(3)) \\ & S_7=(7 / 2)(4+6(3)) \\ & S_7=(7 / 2)(4+18) \\ & S_7=(7 / 2)(22) \\ & S_7=77 \end{aligned}

Therefore, the value of the 7th term $\left(a_7\right)$ is 20 and the sum of the first 7 terms $\left(S_7\right)$ is 77.

Applications of arithmetic sequences:
Arithmetic sequences are commonly used in real-world scenarios, such as:

Calculating compound interest over time.
Predicting population growth or decline.
Estimating the number of terms required to reach a specific value in a series.

February half term is a critical time for A-Level Maths students to focus on revision. With the summer exams fast approaching, this is an opportune moment to consolidate any gaps in knowledge highlighted by recent mock exams. It is crucial to begin preparing a well-structured revision plan to ensure a thorough and systematic review of all topics.

Utilising this time effectively will allow students to strengthen their understanding, address any weaknesses, and enhance their problem-solving skills. By dedicating ample time and effort to revision during February, students can position themselves for success in the upcoming A-Level Maths examinations.

Arithmetic Sequences – Example 1

Solution

\begin{aligned} S_n=a & +(a+d)+(a+2 d)+\ldots \\ & +(a+(n-2) d)+(a+(n-1) d) \end{aligned}

\begin{aligned} S_n=(a & +(n-1) d)+(a+(n-2) d)+\ldots \\ & +(a+2 d)+(a+d)+a \end{aligned}

Adding these gives:

\begin{aligned} & 2 \times S_n=n(2 a+(n-1) d) \\ & S_n=\frac{n}{2}(2 a+(n-1) d) \end{aligned}

Arithmetic Sequences – Example 2

Solution

\begin{aligned} & a=32, d=-5 \\ & S_{50}=\frac{50}{2}(2(32)+(50-1)(-5)) \\ & S_{50}=-4525 \end{aligned}

Arithmetic Sequences – Example 3

Solution

4+9+14+19+\ldots>2000

Using

S_n=\frac{n}{2}(2 a+(n-1) d)

\begin{aligned} 2000 & =\frac{n}{2}(2 \times 4+(n-1) 5) \\ 4000 & =n(8+5 n-5) \\ 4000 & =n(5 n+3) \\ 4000 & =5 n^2+3 n \\ 0 & =5 n^2+3 n-4000 \\ n & =\frac{-3 \pm \sqrt{9+80000}}{10} \\ n & =28.0 \text { or }-28.6 \end{aligned}

So there are 28 items.

Arithmetic Sequences – Example 4

Solution

Substitute $n=20, a=3, \text { and } S_n=288$ into $S_n=\frac{1}{2} n[2 a+(n-1) d]$

\begin{aligned} 288 & =\frac{1}{2} \times 20[(2 \times 3)+19 d] \\ & =10[6+19 d] \\ & =60+190 d \\ 190 d & =228 \\ d & =1.2 \end{aligned}

Arithmetic Sequences – Example 5

\begin{aligned} &\begin{aligned} & a_k=a+(k-1) d \\ & 24=a+(5-1) d \\ & 24=a+4 d \\ & 4=a+(9-1) d \\ & 4=a+8 d \end{aligned}\\ &\begin{aligned} 24 & =a+4 d \\ 4 & =a+8 d \end{aligned} \end{aligned}

20=-4 d \text {, so } d=-5

and a = 44

And for the last term:

\begin{aligned} -36 & =44+(k-1) \times(-5) \\ -80 & =-5(k-1) \\ 16 & =k-1 \\ k & =17 \end{aligned}

Arithmetic Sequences – Example 6

Solution

\begin{aligned} & 616=\frac{1}{2} n[2 \times 5+(n-1) \times 6] \\ & 616=\frac{1}{2} n(10+6 n-6) \\ & 616=\frac{1}{2} n(4+6 n) \\ & 616=2 n+3 n^2 \\ & 3 n^2+2 n-616=0 \\ & (3 n+44)(n-14)=0 \end{aligned}

And n = 14.

In conclusion, an arithmetic sequence in A Level Maths is a sequence of numbers with a constant difference between consecutive terms. It can be used to find specific terms in the sequence and calculate the sum of a given number of terms. Additionally, arithmetic sequences have practical applications in various fields.

If you, or your parents would like to find out more, please just get in touch via email at info@exam.tips or call us on 0800 689 1272.

LAST NAME

POSTCODE

Mobile Number

EMAIL ADDRESS

Message (Please include exam board, school year any specific topics or other relevant details)

Arithmetic Sequences For A Level Maths