Arithmetico-geometric sequence

Not to be confused with Arithmetic–geometric mean.

In mathematics, an arithmetico-geometric sequence is the result of term-by-term multiplication of a geometric progression with the corresponding terms of an arithmetic progression. 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 sequence. 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 series summation of the infinite terms of this sequence is known as an arithmetico-geometric series, and it has been called Gabriel's staircase. In general,

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

The label of arithmetico-geometric sequence may also be given to different objects combining characteristics of both arithmetic and geometric sequences. For instance, the French notion of arithmetico-geometric sequence refers to sequences that satisfy recurrence relations of the form $u_{n+1}=au_{n}+b$, which combine the recurrence relations $u_{n+1}=u_{n}+b$ for arithmetic sequences and $u_{n+1}=au_{n}$ for geometric sequences. These sequences are therefore solutions to a special class of linear difference equation: inhomogeneous first order linear recurrences with constant coefficients.

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 $r=1/2$.

Partial sums

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

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

by r gives

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

Subtracting rS_n from S_n, dividing both sides by $b$, and using the technique of telescoping series (second equality) and the formula for the sum of a finite geometric series (fifth equality) gives

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

as claimed.

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

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

Similarly, the sum

$S={\dfrac {\color {blue}{0*1/6}}{\color {green}{5/6}}}+{\dfrac {\color {blue}{1*1/6}}{\color {green}{1}}}+{\dfrac {\color {blue}{2*1/6}}{\color {green}{6/5}}}+{\dfrac {\color {blue}{3*1/6}}{\color {green}{(6/5)^{2}}}}+{\dfrac {\color {blue}{4*1/6}}{\color {green}{(6/5)^{3}}}}+{\dfrac {\color {blue}{5*1/6}}{\color {green}{(6/5)^{4}}}}+\cdots$

is the sum of an arithmetico-geometric series defined by $d=1/6$, $a=0$, $b=6/5$, and $r=5/6$, and it converges to 6. This sequence corresponds to the expected number of six-sided dice rolls required to obtain a specific value on a die roll, for instance "5". In general, these series with $d=p$, $a=0$, $b=1/(1-p)$, and $r=(1-p)$ give the expectations of "the number of trials until first success" in Bernoulli processes with "success probability" $p$.

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.