Exercise 4.4 — Question 1

Arithmetic-Geometric Sequences

An arithmetic-geometric sequence is formed by multiplying the corresponding terms of an arithmetic sequence and a geometric sequence.

If $\{a,a+d,a+2d,\dots \}$ is an arithmetic sequence and $\{1,r,r^2,\dots \}$ is a geometric sequence, then their term-by-term product gives the arithmetic-geometric sequence:

$a,(a+d)r,(a+2d)r^2,(a+3d)r^3,\dots$

General Term

The $n$-th term (general term) of an arithmetic-geometric sequence is:

$T_n=[a+(n-1)d]r^{n-1}$

where:

• $a$ = first term of the arithmetic part
• $d$ = common difference of the arithmetic part
• $r$ = common ratio of the geometric part

Sum to $n$ Terms

To find $S_n$, use the subtraction method:

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

Multiply both sides by $r$:

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

Subtracting:

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

$(1-r)S_n=a+d\cdot \frac{r(1-r^{n-1})}{1-r}-[a+(n-1)d]r^n$

$S_n=\frac{a}{1-r}+\frac{dr(1-r^{n-1})}{(1-r{)}^2}-\frac{[a+(n-1)d]r^n}{1-r},r\ne 1$

Sum to Infinity

When $|r|<1$, as $n\to \infty$, the terms $r^n\to 0$ and $r^{n-1}\to 0$, so:

$S_{\infty }=\frac{a}{1-r}+\frac{dr}{(1-r{)}^2}$

Sigma Notation

The arithmetic-geometric series can be written using sigma notation as:

$S_n={\sum }_{k=1}^n[a+(k-1)d]r^{k-1}$

The infinite series is:

$S_{\infty }={\sum }_{k=1}^{\infty }[a+(k-1)d]r^{k-1},|r|<1$

Worked Example

Find the general term and sum to infinity of: $1+2r+3r^2+4r^3+\cdots$ where $|r|<1$.

Solution:

• Arithmetic part: $1,2,3,4,\dots$ so $a=1$, $d=1$
• Geometric part: $1,r,r^2,r^3,\dots$ so common ratio $=r$
• General term: $T_n=n\cdot r^{n-1}$
• Sum to infinity: $S_{\infty }=\frac{1}{1-r}+\frac{r}{(1-r{)}^2}=\frac{1}{(1-r{)}^2}$