A relative geometric-tail error · source reference

Printed question

The difference between the sum of the first $n$ terms of the geometric progression and its sum to infinity is less than 0.4% of the sum to infinity. Assuming the value of $r$ is $\frac34$, find the least value of $n$. [3]

Printed working

$$\begin{aligned}|S_n-S_{\infty}|&<\frac{0.4}{100}S_{\infty}\\\left|\frac{a(1-r^n)}{1-r}-\frac a{1-r}\right|&<0.004\left(\frac a{1-r}\right)\\\left|\frac a{1-r}\right|\left|(1-r^n)-1\right|&<0.004\left(\frac a{1-r}\right).\end{aligned}$$

Note $a>0,r=\frac34$, hence $\frac a{1-r}>0$. Thus $|-(\frac34)^n|<0.004$. Since $(\frac34)^n>0$ for all $n$,

$$\left(\frac34\right)^n<0.004\Rightarrow n\ln\frac34<\ln0.004\Rightarrow n>\frac{\ln0.004}{\ln(3/4)}>19.19.$$

Least $n=20$.

Alternatively, because $S_{\infty}>S_n$ (positive $a,r$),

$$S_{\infty}-S_n<0.004S_{\infty}\Rightarrow\frac a{1-r}-\frac{a(1-r^n)}{1-r}<0.004\frac a{1-r}\Rightarrow(3/4)^n<0.004.$$

Taking logs gives the same threshold and answer.

Teacher notes

$|ab|=|a||b|$; $|-b|=b$ for $b>0$. Change the inequality sign when dividing/multiplying both sides by a negative value; $\ln(\frac34)<0$.

Since $a,r$ are positive, all terms are positive. For $m>n$, $u_1+\cdots+u_m>u_1+\cdots+u_n$, so $S_m>S_n$ and $S_{\infty}>S_n$.

Printed answer key

20

Mathematical review and qualifications

For the intended nonzero positive-sum GP, $(S_\infty-S_n)/S_\infty=(3/4)^n$. Taking logarithms gives $n>\ln(0.004)/\ln(3/4)\approx19.19$. Thus $n=20$. Checking adjacent integers gives $(3/4)^{19}\approx0.00423>0.004$ and $(3/4)^{20}\approx0.00317<0.004$.

The percentage statement implicitly concerns a nonzero positive total; a zero GP cannot have a strict error less than zero.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [2022/RVHS/Promo/3(ii)] Question PDF pages 1; solution PDF pages 1, 2. The original files remain private QA records outside this native content.