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:^[4]

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

Example edit

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 ${\displaystyle r=1/2}$ .

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 edit

Multiplying,^[4]

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

by r, gives

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

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

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

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

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^[4]

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

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 edit

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

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

• 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.