Arithmetic and geometric sequences

There are many interesting sequences, but we will limit ourselves here to only two types and discuss them in more detail.

Definition 1

Consider a sequence aa:

(an)=(a1,a2,a3,a4,...,an,an+1,...)(a_n)=(a_1,a_2, a_3, a_4, ..., a_n, a_{n+1}, ...)
  1. (an)(a_n) is called an arithmetic sequence if always the same number dd is added from one term to the next: a1→+da2→+d...→+dan→+dan+1...\begin{array}{lll} a_1 \xrightarrow[]{+d} a_2 \xrightarrow[]{+d} ... \xrightarrow[]{+d} a_n \xrightarrow[]{+d} a_{n+1} ...\end{array} So the recursive formula is an+1⏟new=an⏟old+d(n=1,2,3,...)\underbrace{a_{n+1}}_{new}=\underbrace{a_n}_{old}+d \quad (n=1,2,3,...) The number dd is called common difference because the difference between successive terms is always dd: d=an+1−and=a_{n+1}-a_n.
  2. (an)(a_n) is called a geometric sequence if always the same factor qq is multiplied from one term to the next: a1→⋅qa2→⋅q...→⋅qan→⋅qan+1...\begin{array}{lll} a_1 \xrightarrow[]{\cdot q} a_2 \xrightarrow[]{\cdot q} ... \xrightarrow[]{\cdot q} a_n \xrightarrow[]{\cdot q} a_{n+1} ...\end{array} So the recursive formula is an+1⏟new=an⏟old⋅q\underbrace{a_{n+1}}_{new}=\underbrace{a_n}_{old}\cdot q The number qq is called the common ratio because the ratio between successive terms is always qq: q=an+1anq=\frac{a_{n+1}}{a_n}.
Example 1

Consider again the four sequences from above. For each of these sequences, determine whether it is an arithmetic or geometric sequence, and then also determine the common difference dd or the common quotient qq.

  1. (an)=(1,3,5,...)(a_n)=(1,3,5,...)

  2. (bn)=(1,0.5,0.25,0.125,...)(b_n)=(1, 0.5, 0.25, 0.125, ...)

  3. (cn)=(1,−1,1,−1,...)(c_n)=(1,-1,1,-1,...)

  4. (dn)=(1,2,4,8,...)(d_n)=(1,2,4,8,...)

Solution
  1. arithmetic, d=a2−a1=3−1=2d=a_2-a_1=3-1=2
  2. geometric, q=a2a1=0.51=0.5q=\frac{a_2}{a_1}=\frac{0.5}{1}=0.5
  3. geometric, q=a2a1=−11=−1q=\frac{a_2}{a_1}=\frac{-1}{1}=-1
  4. geometric, q=a2a1=21=2q=\frac{a_2}{a_1}=\frac{2}{1}=2

We have defined the arithmetic and geometric sequence using the recursive formula. However, we can also express the value of the nn-th term using the explicit formula:

Theorem 1

  1. Given an arithmetic sequence aa with initial value a1a_1 and common difference dd. The nn-th term of the sequence is

    an=a1+(n−1)⋅d(n=1,2,3,...)a_{n}=a_1+(n-1)\cdot d\quad (n=1,2,3,...)

  2. Given a geometric sequence aa with initial value a1a_1 and common quotient qq. The nn-th term of the sequence is

    an=a1⋅qn−1(n=1,2,3,...)a_{n}=a_1\cdot q^{n-1}\quad (n=1,2,3,...)
Proof

This is easy to see. For the arithmetic sequence we have

a2=a1+da3=a2+d=a1+2da4=a3+d=a1+3d...an=a1+(n−1)d\begin{array}{lll} a_2&=&a_1+d\\ a_3&=&a_2+d=a_1+2d\\ a_4&=&a_3+d=a_1+3d\\ ... & &\\ a_n&=& a_1+(n-1)d\\ \end{array}

and for the geometric sequence it is

a2=a1⋅qa3=a2⋅q=a1⋅q2a4=a3⋅q=a1⋅q3...an=a1⋅qn−1\begin{array}{lll} a_2&=&a_1\cdot q\\ a_3&=&a_2\cdot q=a_1\cdot q^2\\ a_4&=&a_3\cdot q=a_1\cdot q^3\\ ... & &\\ a_n&=& a_1\cdot q^{n-1}\\ \end{array}
Example 2

Consider the sequence

(an)=(3,3.3,...)(a_n)=(3,3.3,...)

Determine the 2020-th term if (an)(a_n) is

  1. an arithmetic sequence.

  2. a geometric sequence.

Solution
  1. d=a2−a1=0.1d=a_2-a_1=0.1, thus a20=3+0.3⋅19=8.7‾a_{20}=3+0.3\cdot 19=\underline{8.7}
  2. q=a2a1=1.1q=\frac{a_2}{a_1}=1.1, thus a20=3⋅1.119=18.3477...‾a_{20}=3\cdot 1.1^{19}=\underline{18.3477...}