# Arithmetico–geometric sequence

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In mathematics, an arithmetico–geometric sequence is the result of the term-by-term multiplication of a geometric progression with the corresponding terms of an arithmetic progression. Put more plainly, the nth term of an arithmetico–geometric sequence is the product of the nth term of an arithmetic sequence and the nth term of a geometric one. Arithmetico–geometric sequences arise in various applications, such as the computation of expected values in probability theory. For instance, the sequence

${\displaystyle {\dfrac {\color {blue}{0}}{\color {green}{1}}},\ {\dfrac {\color {blue}{1}}{\color {green}{2}}},\ {\dfrac {\color {blue}{2}}{\color {green}{4}}},\ {\dfrac {\color {blue}{3}}{\color {green}{8}}},\ {\dfrac {\color {blue}{4}}{\color {green}{16}}},\ {\dfrac {\color {blue}{5}}{\color {green}{32}}},\cdots }$

is an arithmetico–geometric sequence. The arithmetic component appears in the numerator (in blue), and the geometric one in the denominator (in green).

The denomination may also be applied to different objects presenting characteristics of both arithmetic and geometric sequences; for instance the French notion of arithmetico–geometric sequence refers to sequences of the form ${\displaystyle u_{n+1}=au_{n}+b}$, which generalise both arithmetic and geometric sequences. Such sequences are a special case of linear difference equations.

## Terms of the sequence

The first few terms of an arithmetico–geometric sequence composed of an arithmetic progression (in blue) with difference ${\displaystyle d}$ and initial value ${\displaystyle a}$ and a geometric progression (in green) with initial value ${\displaystyle b}$ and common ratio ${\displaystyle r}$ are given by:[1]

{\displaystyle {\begin{aligned}t_{1}&=\color {blue}a\color {green}b\\t_{2}&=\color {blue}(a+d)\color {green}br\\t_{3}&=\color {blue}(a+2d)\color {green}br^{2}\\&\ \,\vdots \\t_{n}&=\color {blue}[a+(n-1)d]\color {green}br^{n-1}\end{aligned}}}

### Example

For instance, the sequence

${\displaystyle {\dfrac {\color {blue}{0}}{\color {green}{1}}},\ {\dfrac {\color {blue}{1}}{\color {green}{2}}},\ {\dfrac {\color {blue}{2}}{\color {green}{4}}},\ {\dfrac {\color {blue}{3}}{\color {green}{8}}},\ {\dfrac {\color {blue}{4}}{\color {green}{16}}},\ {\dfrac {\color {blue}{5}}{\color {green}{32}}},\cdots }$

is defined by ${\displaystyle d=b=1}$, ${\displaystyle a=0}$, and ${\displaystyle r={\frac {1}{2}}}$.

## Sum of the terms

The sum of the first n terms of an arithmetico–geometric sequence has the form

{\displaystyle {\begin{aligned}S_{n}&=\sum _{k=1}^{n}t_{k}=\sum _{k=1}^{n}\left[a+(k-1)d\right]br^{k-1}\\&=ab+[a+d]br+[a+2d]br^{2}+\cdots +[a+(n-1)d]br^{n-1}\\&=A_{1}G_{1}+A_{2}G_{2}+A_{3}G_{3}+\cdots +A_{n}G_{n},\end{aligned}}}

where ${\displaystyle A_{i}}$ and ${\displaystyle G_{i}}$ are the ith terms of the arithmetic and the geometric sequence, respectively.

This sum has the closed-form expression

{\displaystyle {\begin{aligned}S_{n}&={\frac {ab-(a+nd)\,br^{n}}{1-r}}+{\frac {dbr\,(1-r^{n})}{(1-r)^{2}}}\\&={\frac {A_{1}G_{1}-A_{n+1}G_{n+1}}{1-r}}+{\frac {dr}{(1-r)^{2}}}\,(G_{1}-G_{n+1}).\end{aligned}}}

### Proof

Multiplying,[1]

${\displaystyle S_{n}=ab+[a+d]br+[a+2d]br^{2}+\cdots +[a+(n-1)d]br^{n-1}}$

by r, gives

${\displaystyle rS_{n}=abr+[a+d]br^{2}+[a+2d]br^{3}+\cdots +[a+(n-1)d]br^{n}.}$

Subtracting rSn from Sn, and using the technique of telescoping series gives

{\displaystyle {\begin{aligned}(1-r)S_{n}={}&\left[ab+(a+d)br+(a+2d)br^{2}+\cdots +[a+(n-1)d]br^{n-1}\right]\\[5pt]&{}-\left[abr+(a+d)br^{2}+(a+2d)br^{3}+\cdots +[a+(n-1)d]br^{n}\right]\\[5pt]={}&ab+db\left(r+r^{2}+\cdots +r^{n-1}\right)-\left[a+(n-1)d\right]br^{n}\\[5pt]={}&ab+db\left(r+r^{2}+\cdots +r^{n-1}+r^{n}\right)-\left(a+nd\right)br^{n}\\[5pt]={}&ab+dbr\left(1+r+r^{2}+\cdots +r^{n-1}\right)-\left(a+nd\right)br^{n}\\[5pt]={}&ab+{\frac {dbr(1-r^{n})}{1-r}}-(a+nd)br^{n},\end{aligned}}}

where the last equality results of the expression for the sum of a geometric series. Finally dividing through by 1 − r gives the result.

## Infinite series

If −1 < r < 1, then the sum S of the arithmetico–geometric series, that is to say, the sum of all the infinitely many terms of the progression, is given by[1]

{\displaystyle {\begin{aligned}S&=\sum _{k=1}^{\infty }t_{k}=\lim _{n\to \infty }S_{n}\\&={\frac {ab}{1-r}}+{\frac {dbr}{(1-r)^{2}}}\\&={\frac {A_{1}G_{1}}{1-r}}+{\frac {dG_{1}r}{(1-r)^{2}}}.\end{aligned}}}

If r is outside of the above range, the series either

• diverges (when r > 1, or when r = 1 where the series is arithmetic and a and d are not both zero; if both a and d are zero in the later case, all terms of the series are zero and the series is constant)
• or alternates (when r ≤ −1).

### Example: application to expected values

For instance, the sum

${\displaystyle S={\dfrac {\color {blue}{0}}{\color {green}{1}}}+{\dfrac {\color {blue}{1}}{\color {green}{2}}}+{\dfrac {\color {blue}{2}}{\color {green}{4}}}+{\dfrac {\color {blue}{3}}{\color {green}{8}}}+{\dfrac {\color {blue}{4}}{\color {green}{16}}}+{\dfrac {\color {blue}{5}}{\color {green}{32}}}+\cdots }$,

being the sum of an arithmetico–geometric series defined by ${\displaystyle d=b=1}$, ${\displaystyle a=0}$, and ${\displaystyle r={\frac {1}{2}}}$, converges to ${\displaystyle S=2}$.

This sequence corresponds to the expected number of coin tosses before obtaining "tails". The probability ${\displaystyle T_{k}}$ of obtaining tails for the first time at the kth toss is as follows:

${\displaystyle T_{1}={\frac {1}{2}},\ T_{2}={\frac {1}{4}},\dots ,T_{k}={\frac {1}{2^{k}}}}$.

Therefore, the expected number of tosses is given by

${\displaystyle \sum _{k=1}^{\infty }kT_{k}=\sum _{k=1}^{\infty }{\frac {\color {blue}k}{\color {green}2^{k}}}=S=2}$ .

## References

1. ^ a b c K. F. Riley; M. P. Hobson; S. J. Bence (2010). Mathematical methods for physics and engineering (3rd ed.). Cambridge University Press. p. 118. ISBN 978-0-521-86153-3.