A variable geometric ratio · source reference

Printed question

The geometric series is given by $\frac{3x-2}{4x}+\frac{(3x-2)^2}{4x}+\frac{(3x-2)^3}{4x}+\cdots$ where $x\ne0$. Explain why the sum to infinity exists for $\frac13<x<1$, and find its value in terms of $x$. [5]

Printed working

The common ratio is $r=3x-2$.

$$\frac13<x<1\Rightarrow1<3x<3\Rightarrow-1<3x-2<1\Rightarrow|3x-2|<1\Rightarrow|r|<1.$$

Hence the sum to infinity exists for $\frac13<x<1$.

$$S_{\infty}=\frac{(3x-2)/(4x)}{1-(3x-2)}=\frac{3x-2}{12x(1-x)}.$$

Teacher notes

GP: the sum to infinity exists when $|r|<1$, i.e. $-1<r<1$, where $r$ is the common ratio.

Sum to infinity $=\frac a{1-r}$, where $a$ is the first term and $r$ is the common ratio.

Printed answer key

$\frac{3x-2}{12x(1-x)}$

Mathematical review and qualifications

$|3x-2|<1$ is equivalent to $1/3<x<1$. On this interval the sum is $\frac{(3x-2)/(4x)}{1-(3x-2)}=\frac{3x-2}{12x(1-x)}$. At $x=2/3$ all terms are zero, and the same expression correctly gives zero.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [2020/DHS/Prelim/I/1] Question PDF pages 1; solution PDF pages 1. The original files remain private QA records outside this native content.