Arithmetico-geometric sequence

Not to be confused with Arithmetic–geometric mean.

In mathematics, arithmetico-geometric sequence is the result of term-by-term multiplication of a geometric progression with the corresponding terms of an arithmetic progression. Put 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

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

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

The summation of this infinite sequence is known as an arithmetico-geometric series, and its most basic form has been called Gabriel's staircase:

$\sum _{k=1}^{\infty }{\color {blue}k}{\color {green}r^{k}}={\frac {r}{(1-r)^{2}}},\quad \mathrm {for\ } 0<r<1$

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 $u_{n+1}=au_{n}+b$, 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 $d$ and initial value $a$ and a geometric progression (in green) with initial value $b$ and common ratio $r$ are given by:

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

Example

For instance, the sequence

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

is defined by $d=b=1$, $a=0$, and $r=1/2$.

Sum of the terms

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

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

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

This sum has the closed-form expression

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

Proof

Multiplying,

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

by r, gives

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

Subtracting rS_n from S_n, and using the technique of telescoping series gives

${\begin{aligned}(1-r)S_{n}={}&\left[a+(a+d)r+(a+2d)r^{2}+\cdots +[a+(n-1)d]r^{n-1}\right]\\[5pt]&{}-\left[ar+(a+d)r^{2}+(a+2d)r^{3}+\cdots +[a+(n-1)d]r^{n}\right]\\[5pt]={}&a+d\left(r+r^{2}+\cdots +r^{n-1}\right)-\left[a+(n-1)d\right]r^{n}\\[5pt]={}&a+d\left(r+r^{2}+\cdots +r^{n-1}+r^{n}\right)-\left(a+nd\right)r^{n}\\[5pt]={}&a+dr\left(1+r+r^{2}+\cdots +r^{n-1}\right)-\left(a+nd\right)r^{n}\\[5pt]={}&a+{\frac {dr(1-r^{n})}{1-r}}-(a+nd)r^{n},\end{aligned}}$

where the last equality follows from the formula for the sum of a finite geometric series. Finally, rearranging to group the first and third terms of that expression, rewriting each group in terms of ${\textstyle A_{i}}$ and ${\textstyle G_{i},}$ and then dividing through by 1 − r gives the closed-form expression claimed above.

Infinite series

If −1 < r < 1, then the sum S of the arithmetico-geometric series, that is to say, the limit of the partial sums of all the infinitely many terms of the progression, is given by

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

If r is outside of the above range, b is not zero, and a and d are not both zero, the series diverges.

Example: application to expected values

The sum

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

is the sum of an arithmetico-geometric series defined by $d=b=1$, $a=0$, and $r={\frac {1}{2}}$, and it converges to $S=2$. This sequence corresponds to the expected number of coin tosses required to obtain "tails". The probability $T_{k}$ of obtaining tails for the first time at the kth toss is as follows:

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

Therefore, the expected number of tosses to reach the first "tails" is given by

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

Further reading

• D. Khattar. The Pearson Guide to Mathematics for the IIT-JEE, 2/e (New ed.). Pearson Education India. p. 10.8. ISBN 81-317-2876-5.
• P. Gupta. Comprehensive Mathematics XI. Laxmi Publications. p. 380. ISBN 81-7008-597-7.