Arithmetic and geometric series

Definition 1

Given a sequence

(a_n)=(a_1, a_2, ...)

The corresponding series

(s_n)=(s_1, s_2, ...)

is called

an arithmetic series, if $(a_n)$ is an arithmetic sequence.
a geometric series, if $(a_n)$ is a geometric sequence.

Theorem 1: Arithmetic and geometric sum formulas

Proof

Here are the proofs. Given the sequence

(a_n)=(a_1,a_2,...,a_n, ...)

and

s_n=a_1+a_2+...+a_n

Arithmetic summation formula

Let $(a_n)$ be an arithmetic sequence with common difference $d$ :

\begin{array}{lll} a_1 \xrightarrow[]{+d} a_2 \xrightarrow[]{+d} a_3 \xrightarrow[]{+d} ... \xrightarrow[]{+d} a_{n-2} \xrightarrow[]{+d} a_{n-1} \xrightarrow[]{+d} a_n \, ...\end{array}

We now first calculate the twofold of $s_n$ , and order the terms as follows:

\begin{array}{lll} 2s_n &=& a_1 &+& a_2 &+& a_3 & + & ... & + & a_{n-2} &+&a_{n-1} &+ & a_n \\ & + & a_{n} &+& a_{n-1} &+& a_{n-2} & + & ... & + & a_3 &+&a_2 &+ & a_1 \end{array}

Now notice that

a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2} = ...

Indeed, we have

\begin{array}{lll} a_2+a_{n-1} & = &\underbrace{a_1+d}_{a_2}+\underbrace{a_n-d}_{a_{n-1}}\\ & = & a_1+a_n\\ a_3+a_{n-2} & = &\underbrace{a_1+2d}_{a_3}+\underbrace{a_n-2d}_{a_{n-2}}\\ & = & a_1+a_n\\ ...&&\\ \end{array}

So we see that we add $n$ times the term $a_1+a_n$ .

2s_n = n(a_1+a_n)

Thus

s_n=\frac{n(a_1+a_n)}{2}

Geometric summation formula

Let $(a_n)$ be a geometric sequence with common quotient $q$ :

\begin{array}{lll} a_1 \xrightarrow[]{\cdot q} a_2 \xrightarrow[]{\cdot q} a_3 \xrightarrow[]{\cdot q} ... \xrightarrow[]{\cdot q} a_{n-1}\xrightarrow[]{\cdot q} a_{n}\, ...\end{array}

It is

\begin{array}{lll} s_n &=& a_1+a_2+a_3+...+a_{n-1}+a_n\\ &=& a_1 + a_1 q +a_1 q^2 +...+ a_1 q^{n-2} + a_1 q^{n-1} \end{array}

To determine $s_n$ , we take a diversion, and first determine the sum $s_n-q \cdot s_n$ :

\begin{array}{lll} s_n-q\cdot s_n &=& a_1+a_1 q+a_1 q^2+...+a_1 q^{n-2}+a_1 q^{n-1}\\ && -a_1 q - a_1 q^2- a_1 q^3 -...-a_1 q^{n-1}-a_1 q^{n}\\ &=& a_1-a_1 q^n \end{array}

We get

\begin{array}{lll} s_n-q\cdot s_n &=& a_1-a_1 q^n\\ s_n(1-q) &=& a_1(1-q^n)\\ s_n =a_1\frac{1-q^n}{1-q} \end{array}

The n-th term of an arithmetic series is $s_n=a_1+...+a_n=\frac{n(a_1+a_n)}{2}$
The n-th term of a geometric series is $s_n=a_1+...+a_n=a_1\cdot \frac{1-q^n}{1-q}$ where $q$ is the common ratio of the geometric sequence $(a_n)$ .

Example 1

Determine the sum of the first

$45$ terms of the sequence $(a_n)=(2,5,8,...)$ .
$20$ terms of the sequence $(a_n)=(1,2,3,4, ...)$ .
$10$ terms of the sequence $(a_n)=(2,6,18,54,...)$ .
$12$ terms of the sequence $(a_n)=(-\frac{1}{4},\frac{1}{6},-\frac{1}{9}, \frac{2}{27},...)$ .

Solution

arithmetic sequence with $d=3$ . So $s_{45}=\frac{45\cdot(2+a_{45})}{2}$ and because of $a_{45}=2+44\cdot 3=134$ ist $s_{45}=\underline{3060}$ .
arithmetic sequence with $d=1$ and $a_{20}=20$ . So $s_{20}=\frac{20\cdot 21}{2}=\underline{210}$
geometric sequence with $q=3$ . So $s_{10}=2\cdot \frac{1-3^{10}}{1-3}=\underline{59048}$
geometric sequence with $q=-2/3$ . So $s_{12}=-\frac{1}{4}\cdot \frac{1-\left(-\frac{2}{3}\right)^{12}}{1-\left(-\frac{2}{3}\right)}=\underline{-0.1488...}$

Exercise 1

Find the next two terms of the arithmetic sequence $(a_n)=(-0.25, 0.32, ...)$ . What is the sum of the first $20$ terms?
Find the next two terms of the geometric sequence $(a_n)=(3, -4, ...)$ . What is the sum of the first $20$ terms?
An arithmetic sequence has the value $16.5$ as the $7$ th term and the value $24$ as the $12$ th term. Find the value of the $26$ th term.
The $10$ -th element of a geometric sequence has the value $2$ and the $13$ -th element has the value $1.1$ . Find the value of the $17$ th term.
A car has the value $25600.-$ when new (month one). Every month it loses $90.-$ in value. When will the value of the car fall below $15000.-$ for the first time?
A computer business expects to sell $20$ of computers in the first month, $23$ in the second month, $26$ in the third month, and so on. How many months will it take for the business to sell a total of $1000$ of computers?
The new value of a car is $340\,000.-$ (year one). It loses $15\%$ of its value from the previous year every year. After how many years does the value of the car fall below $100\,000.-$ for the first time?
Draw the terms of the arithmetic sequence $(a_n)=(1,1.5,2,...)$ in a coordinate system as points ( $n$ along the $x$ -axis, $a_n$ along the $y$ -axis). Determine the function equation of the graph that passes through these points.
Draw the terms of the geometric sequence $(a_n)=(1,1.5,2.25,...)$ as points in a coordinate system. Determine the function equation of the graph that passes through these points.

Solution

$d=0.57$ , $a_3=0.32+0.57=\underline{0.89}$ , $a_4=0.89+0.57=\underline{1.46}$ . The sum is $s_{20}=\frac{20\cdot (-0.25+a_{20})}{2}$ and with $a_{20}=-0.25+19\cdot 0.57 = 10.58$ we get $s_{20}=\underline{103.3}$ .
$q=a_2/a_1=-4/3=-\frac{4}{3}$ , $a_1=3$ , so $a_3=a_2\cdot q = -4 \cdot (-4/3)=\underline{16/3}$ , $a_4=a_3\cdot q = (16/3)\cdot(-4/3)=\underline{-64/9}$ . The sum is $s_{20}=3\cdot \frac{1-(-\frac{4}{3})^{20}}{1-(-\frac{4}{3})}=\underline{-404.147...}$
$a_7=16.5 \xrightarrow[]{+d} a_8 \xrightarrow[]{+d} a_9 \xrightarrow[]{+d} a_{10} \xrightarrow[]{+d} a_{11} \xrightarrow[]{+d} a_{12}=24$ Thus $16.5+5\cdot d = 24$ and $d=(24-16.5)/5=1.5$ . So it is $a_{26}=a_{12}+14\cdot 1.5= \underline{45}$ .
$a_{10} = 2 \xrightarrow[]{\cdot q} a_{11} \xrightarrow[]{\cdot q} a_{12} \xrightarrow[]{\cdot q} a_{13}= 1.1$ Thus $2\cdot q^3 =1.1$ , so $q = (\frac{1.1}{2})^{1/3}=0.55^{1/3}$ . It follows $a_{17}=a_{13}\cdot q^4 = 1.1 \cdot (0.55)^{4/3}=\underline{0.495...}$ .
$a_1=25600$ , $d=-90$ . Find $n$ such that $a_n=a_1+(n-1)d<15000$ , so $256000-90(n-1)<15000$ . We solve the equation $25600-90(n-1)=15000$ , so $n-1=(25600-15000)/90= 117.7$ and hence $n=118.7$ . So it happens in month $\underline{119}$ (so after $\underline{118}$ months).
$(a_n)=20,23,26,...$ , so $d=3$ and $a_n=20+(n-1)\cdot 3$ . Find $n$ with $s_n=a_1+...+a_n=1000$ , so we have to solve the equation $\frac{n(20+a_n)}{2}=1000$ . It follows $\frac{n(20+20+(n-1)\cdot 3)}{2}=\frac{n(37+3n)}{2}=1000$ , and so $3n^2 + 37n-2000=0$ . Using the midnight formula, we get $n_1=20.37$ and $n_2=-32.7$ . So it happens in month $21$ , so after $\underline{20}$ months.
$a_1 \xrightarrow[]{-15\%} a_2 \xrightarrow[]{-15\%} a_3 \xrightarrow[]{-15\%} ...$ Geometric sequence with $a_1=340000$ , $a_2=340000-0.15\cdot 340000=289000$ , and $a_3=289000-0.15\cdot 289000=245650$ . So we have $q=a_2/a_1=a_3/a_2=0.85$ . Find $n$ -th term with $a_n=340000\cdot 0.85^{n-1}=100000$ . It follows $0.85^{n-1}=\frac{100000}{340000}=0.294...$ and thus $n=\frac{\ln(0.294...)}{\ln(0.85)}+1=8.5$ (apply logarithm), i.e. at year $9$ , so after $\underline{8}$ years.
This is typical linear growth, which has been discussed before: $\begin{array}{lllcc} y:&& 1 \xrightarrow[]{+0.5} &1.5& \xrightarrow[]{+0.5} &2 &\xrightarrow[]{+0.5}\, ... \xrightarrow[]{+0.5} y\\ x: && 1 \xrightarrow[]{+1} &2 &\xrightarrow[]{+1} &3& \xrightarrow[]{+1}\,... \xrightarrow[]{+1} x\\ \end{array}$ So we get the linear function $f(x)=\underline{0.5x+0.5}$ The graph of this function passes through the points of the sequence.
This is typical exponential growth, which has been discussed before: $\begin{array}{lllcc} y:&& 1 \xrightarrow[]{\cdot 1.5} &1.5& \xrightarrow[]{\cdot 1.5} &2.25 &\xrightarrow[]{\cdot 1.5} \, ...\xrightarrow[]{\cdot 1.5} y\\ x: && 1 \xrightarrow[]{+1} &2 &\xrightarrow[]{+1} &3& \xrightarrow[]{+1}\,...\xrightarrow[]{\cdot 1.5} x\\ \end{array}$ So we get the exponential function $f(x)=\underline{1.5^{x-1}}$ The graph of this function passes through the points of the sequence.