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Linear (Arithmetic) Sequences Simplified Revision Notes

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Linear (Arithmetic) Sequences

Introduction

infoNote

A sequence is a set of objects (usually numbers) listed in order.

infoNote

An arithmetic sequence is a sequence of numbers such that the difference between any two consecutive terms is constant. This constant difference is called the common difference, denoted as dd.

The first term of any arithmetic is denoted as aa, so in general, an arithmetic sequence is :

a,a+d,a+2n,a+3d,a+4d,...a,a+d,a+2n, a+3d,a+4d, ...

Each positional terms is denoted as TnT_{n} where nn is the position of the term. So T1=a,T2=a+dT_1=a, T_2=a+d and so on.

The general term of any arithmetic sequence is

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General term of an arithmetic sequence : Page 22

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

Example

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List the first 55 terms of arithmetic sequence with a=6a=6 and d=2d=-2.

  • T1=a+(11)(2)=a=6T_1=a+(1-1)(-2)=a=6
  • T2=6+(21)(2)=4T_2=6+(2-1)(-2)=4
  • T3=6+(31)(2)=2T_3=6+(3-1)(-2)=2
  • T4=6+(41)(2)=0T_4=6+(4-1)(-2)=0
  • T5=6+(51)(2)=2T_5=6+(5-1)(-2)=-2

Example

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If 6,p,q,326,p, q,32 are the first four consecutive terms in an arithmetic sequence, find pp and qq, and then find the general term.

  • T1=a+(11)(d)=a=6T_1=a+(1-1)(d)=a=6
  • T4=6+(41)(d)=32T_4=6+(4-1)(d)=32
T4=6+3d=323d=26d=263\begin{align*} T_4=6+3d&=32 \\ 3d&=26 \\ d&=\tfrac{26}{3} \end{align*}

Once aa and dd are known, we can find any term, TnT_n.

  • p=T2=(6+263)=443p=T_2=(6+\tfrac{26}{3})=\tfrac{44}{3}

  • q=T3=(443+263)=703q=T_3=(\tfrac{44}{3}+\tfrac{26}{3})=\tfrac{70}{3} The general term :

  • Tn=a+(n1)d=6+(n1)263T_n=a+(n-1)d=6+(n-1)\tfrac{26}{3}

Example

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Which term is 249249 in the arithmetic sequence 2,15,28,...2,15,28,... ?

First let's determine the common difference, which can be found by taking any term subtracted by the term before it, in general :

d=TnTn1d=T_n-T_{n-1}
  • T3T2=2815=13=dT_3-T_2=28-15=13=d

Now we are looking for the position of the term 249249, so nn is the unknown variable :

Let Tn=249T_n=249. Use the general term formula :

Tn=a+(n1)d249=2+(n1)(13)247=13n13260=13n20=n\begin{align*} T_n&=a+(n-1)d \\ 249&=2+(n-1)(13) \\ 247&=13n-13 \\ 260&=13n \\ 20&=n \end{align*}

nn must be a natural number, so getting anything that isn't a natural number indicates a mistake.

:::

Proving a Sequence is Arithmetic

To show that any sequence is arithmetic, we need to prove that the common difference is constant, in general, if a sequence is arithmetic then :

d=TnTn1=cd=T_n-T_{n-1}=c

where cc is some constant.

Example

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Determine if the general term Tn=n2+3nT_n=n^2+3n forms an arithmetic sequence.

We need TnT_n (given to us already) and Tn1T_{n-1}, so we can just sub in n1n-1 for nn.

Tn=n2+3nTn1=(n1)2+3(n1)=n22n+1+3n3=n2+n2\begin{align*} T_n&=n^2+3n \\ T_{n-1}&=(n-1)^2+3(n-1) \\ &=n^2-2n+1+3n-3 \\ &=n^2+n-2 \end{align*}

Now determine if TnTn1T_n-T_{n-1} is constant :

TnTn1=(n2+3n)(n2+n2)=n2+3nn2n+2=2n+2\begin{align*} T_n-T_{n-1}&=(n^2+3n)-(n^2+n-2) \\ &=n^2+3n-n^2-n+2 \\ &= 2n+2 \end{align*}

2n+22n+2 is not constant, so the sequence formed from TnT_n is not arithmetic.

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