Category Archives: Arithmetic

Deriving the sum formula for an Arithmetic Progression

Arithmetic progressions (or arithmetic sequences) are sequences with a common difference (i.e. the same number is added or subtracted to get the next number in the sequence).

For example,

2, 5, 8, 11, 14, ...

or

12,10, 8, 6, 4, ...

The n^{th} term of an arithmetic progression is T_n=a+(n-1)d where a is the first term and d is the common difference.

i.e. For the sequence above, T_4=12+(4-1)(-2)=6

An arithmetic series is the sum of the arithmetic progression.

For example, if the sequence is

3, 7, 11, 15, ...

then S_1=3, S_2=3+7=10, S_3=3+7+11=21

The series is also a sequence and we are going to find the general term, S_n.

    \begin{equation*}S_n=T_1+T_2+T_3+...+T_n\end{equation}

which we can write as

    \begin{equation*}S_n=a+a+d+a+2d+...+a+(n-1)d\end{equation}

Now, I am going to write that in reverse order (to make the next bit more obvious)

    \begin{equation*}S_n=a+(n-1)d+a+(n-2)d+a+(n-3)d+ ... +a\end{equation}

I am going to add the two versions of S_n together

Each term has an a and there are n terms, so we now have 2na. The d terms, we going to group together

d+(n-1)d+2d+(n-2)d+3d+(n-3)d+ ... +(n-1)d+d

Which simplifies to nd+nd+nd+ ...+nd and we have (n-1) d terms. So this part of the sum is (n-10nd

Thus we have

    \begin{equation*}2S_n=2an+(n-1)nd\end{equation}

Which simplifies to

(1)   \begin{equation*}S_n=\frac{n}{2}(2a+(n-1)d)\end{equation*}

Let’s test it, remember the sequence 3, 7, 11, 15, .... We know S_3=21

    \begin{equation*}S_3=\frac{3}{2}(2(3)+(3-1)(4))=\frac{3}{2}(14)=21\end{equation}

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Filed under Algebra, Arithmetic, Sequences, Year 11 Mathematical Methods

Arithmetic Sequence

I did this question with on of my year 11 students. I think the algebra and the subscripts can be a bit tricky.

If T_m=n and T_n=m, then prove that T_{m+n}=0. Here where T_n and T_m are terms of an arithmetic sequence.
Mathematics Methods Units 1&2 – Exercise 15B Question 19

If T_m=n then,

(1)   \begin{equation*}n=a+(m-1)d\end{equation*}


And if T_n=m then,

(2)   \begin{equation*}m=a+(n-1)d\end{equation*}


Subtract equation (2) from equation (1)

    \begin{equation*}n-m=(m-1)d-((n-1)d)\end{equation*}


    \begin{equation*}n-m=md-nd\end{equation*}


(3)   \begin{equation*}n-m=d(m-n)\end{equation*}


Therefore d must equal -1
Substitute d=-1 into equation (1)

    \begin{equation*}n=a+(m-1)(-1)\end{equation*}


(4)   \begin{equation*}n=a-m+1\end{equation*}


Therefore a=n+m-1


(5)   \begin{equation*}T_{m+n}=a+(m+n-1)d\end{equation*}


Substitute a=n+m-1 and d=-1 into equation (5)

    \begin{equation*}$T_{m+n}=n+m-1+(m+n-1)(-1)$\end{equation*}


    \begin{equation*}$T_{m+n}=n+m-1-m-n+1$\end{equation*}


(6)   \begin{equation*}$T_{m+n}=0$\end{equation*}

As you can see from equation (6), T_{m+n}=0

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Filed under Algebra, Arithmetic, Sequences, Year 11 Mathematical Methods