Odd-sized groups of geometric terms · source reference

Printed question

The sum of the first $n$ terms, $S_n$, is $S_n=\frac1a[1-(a-1)^n]$, where $a$ is a constant, $a\ne1$, $n\in\mathbb Z^+$. Obtain the $n$th term $T_n$ and prove that “$S_n$ is a geometric series” (printed wording). [3]

The sequence $\{T_n\}$ is grouped as $(T_1),(T_2,T_3,T_4),(T_5,T_6,T_7,T_8,T_9),\ldots$, each subsequent bracket having 2 more terms than the previous. Find:

(i) total terms in the first $n$ brackets [2];

(ii) the middle term of bracket 11 in terms of $a$ [2];

(iii) the range of $a$ for the sum to infinity to exist; and the least $n$ for the sum of all terms in the first $n$ brackets to be within 0.1% of the sum to infinity when $a=\frac{39}{20}$. [4]

Printed working

Given $S_n=\frac1a[1-(a-1)^n]$,

$$\begin{aligned}T_n&=S_n-S_{n-1}\quad(n\ge2)\\&=\frac1a[1-(a-1)^n]-\frac1a[1-(a-1)^{n-1}]\\&=-\frac1a(a-1)^n+\frac1a(a-1)^{n-1}\\&=\frac1a(a-1)^{n-1}(-a+1+1)\\&=\frac1a(a-1)^{n-1}(2-a),\quad n\in\mathbb Z^+.\end{aligned}$$ $$\frac{T_n}{T_{n-1}}=\frac{a^{-1}(a-1)^{n-1}(2-a)}{a^{-1}(a-1)^{n-2}(2-a)}=a-1,$$

constant independent of $n$. The printed conclusion says “$S_n$ is a geometric series”; it is the underlying $T_n$ terms that form the GP. Common ratio $a-1$.

(i) Terms through $n$ brackets:

$$1+3+5+\cdots=\frac n2[2(1)+(n-1)2]=n^2.$$

(ii) Bracket 11 has $1+(11-1)2=21$ terms; its middle is the 11th. Terms through bracket 10: $10^2=100$. Thus the middle is $T_{111}=\frac1a(a-1)^{110}(2-a)$.

(iii) Infinite sum requires $|a-1|<1\Rightarrow-1<a-1<1\Rightarrow0<a<2$.

At $a=39/20$, $r=a-1=19/20<1$.

$$S_{\infty}=\lim_{n\to\infty}\frac1a[1-(a-1)^n]=\frac1a[1-0]=\frac1a=\frac{20}{39}.$$

Alternative: $S_{\infty}=\frac{(2-a)/a}{1-19/20}=\frac{1/39}{1/20}=20/39$.

Within 0.1% means $|S_{n^2}-S_{\infty}|<0.1\%S_{\infty}$:

$$\left|\frac1a[1-(a-1)^{n^2}]-\frac1a\right|<0.001\frac1a\Rightarrow(a-1)^{n^2}<0.001.$$ $$\left(\frac{19}{20}\right)^{n^2}<0.001\Rightarrow n^2>\frac{\lg0.001}{\lg(19/20)}.$$

Since $n>0$, $n>11.605$. Least $n=12$.

Teacher notes

$S_n=T_1+\cdots+T_n$, $S_{n-1}=T_1+\cdots+T_{n-1}$, so $T_n=S_n-S_{n-1}$.

For a GP, prove $T_n/T_{n-1}$ or $T_{n+1}/T_n$ constant using general terms; the constant is its common ratio.

Bracket sizes form an AP with $a=1,d=2$: sum $\frac n2[2a+(n-1)d]$, term $a+(n-1)d$.

For convergence, $|r|<1$, $r=a-1$. Note $S_1=T_1$.

Since $a=39/20$, $1/a>0,a-1>0$; the absolute error simplifies to $a^{-1}(a-1)^{n^2}$.

Printed answer key

$T_n=\frac1a(a-1)^{n-1}(2-a)$, $n\in\mathbb Z^+$; (i) $n^2$; (ii) $\frac1a(a-1)^{110}(2-a)$; (iii) $0<a<2$; 12

Mathematical review and qualifications

The formula requires $a \ne0$. $T_n=\frac{2-a}{a}(a-1)^{n-1}$, so the underlying terms form a GP of ratio $a-1$. The printed request says $S_n$ is a geometric series; the intended statement concerns these terms.

(i) $1+3+\cdots+(2n-1)=n^2$. (ii) Bracket 11 contains indices 101 through 121; its middle is $T_{111}=\frac{2-a}{a}(a-1)^{110}$.

(iii) For the nonzero GP, $|a-1|<1$ gives $0<a<2$ (with the stated exclusion $a \ne1$). At $a=39/20$, ratio is $19/20$. The relative tail after $n^2$ terms is $(19/20)^{n^2}<0.001$, requiring $n^2>\ln(0.001)/\ln(19/20)\approx134.67$. Thus least $n=12$.

Qualification: $a=2$ makes every term zero and the printed partial sums zero, so the series converges despite ratio 1. The literal convergence set also includes this degenerate endpoint; $a=0$ remains undefined.

The source excludes a=1 but omits the necessary a≠0 restriction; its convergence answer also omits the zero-series endpoint a=2.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [modified 2011/AJC/Prelim/I/12] Question PDF pages 3; solution PDF pages 12, 13. The original files remain private QA records outside this native content.