Half the infinite sum and alternate indices · source reference

Printed question

(a) In a convergent geometric progression, the sum of the first $n$ terms equals the sum of the remaining terms. Given also that the $(n+1)$th term is $\frac12$, determine the $(2n+1)$th term. [4]

(b) An arithmetic progression has first term $a$ and common difference $d$. The eighth, third and second terms are successive terms of an infinite geometric progression. If the first GP term is 10, find the sum of its even-numbered terms. [4]

A sequence is formed whose $n$th term is $\ln|u_nu_{n+1}|$, where $u_n$ is the $n$th term of the geometric progression. Show that it forms an AP and state the common difference. [3]

Printed working

(a) $S_n=S_{\infty}-S_n\Rightarrow2S_n=S_{\infty}$.

$$2\frac{a(1-r^n)}{1-r}=\frac a{1-r}\Rightarrow2(1-r^n)=1\Rightarrow r^n=\frac12.$$

Given $ar^n=\frac12$, hence $a=1$.

$$(2n+1)\text{th term}=ar^{2n}=a(r^n)^2=(1/2)^2=\frac14.$$

(b) $(a+7d),(a+2d),(a+d)$ form a convergent GP.

$$r=\frac{a+2d}{a+7d}=\frac{a+d}{a+2d}\Rightarrow3d^2=-4ad\Rightarrow d=-\frac43a.$$

($d\ne0$ since $r\ne1$ for a convergent GP.)

$$r=\frac{a+d}{a+2d}=\frac{a-4a/3}{a-8a/3}=\frac15.$$

Even-numbered sum $=\frac{10/5}{1-(1/5)^2}=\frac{25}{12}$.

Let $t_n=\ln|u_nu_{n+1}|$.

$$\begin{aligned}t_n-t_{n-1}&=\ln|u_nu_{n+1}|-\ln|u_{n-1}u_n|\\&=\ln\left|\frac{u_nu_{n+1}}{u_{n-1}u_n}\right|=\ln\left|\frac{u_{n+1}}{u_{n-1}}\right|\\&=\ln\left|\frac{10r^n}{10r^{n-2}}\right|=\ln(1/5)^2=2\ln(1/5).\end{aligned}$$

It is constant independent of $n$, so AP with common difference $-2\ln5$.

Teacher notes

The full GP sum is $T_1+\cdots+T_n+T_{n+1}+\cdots$; the first bracket is $S_n$, the remaining bracket is the remaining-tail sum.

Even-indexed GP terms $u_2,u_4,u_6,\ldots=ar,ar^3,ar^5,\ldots$ have first term $ar$ and common ratio $r^2$, so sum $ar/(1-r^2)$.

To prove a sequence is AP, use general consecutive terms, proving a constant difference independent of $n$. This constant is the common difference. It is wrong to rely on specific terms only: $u_3-u_2=u_2-u_1$ proves the first three are AP, not later $u_4,u_5,\ldots$.

Printed answer key

(a) $\frac14$; (b) $\frac{25}{12}$; $-2\ln5$

Mathematical review and qualifications

(a) Equality gives $1-r^n=r^n$, so $r^n=1/2$. Since $ar^n=1/2$, $a=1$ and $ar^{2n}=1/4$.

(b) With AP first term $a$ and difference $d$, $(a+2d)^2=(a+7d)(a+d)$ gives $d(4a+3d)=0$. For the intended convergent branch, $d \ne0$, so $d=-4a/3$ and $r=(a+d)/(a+2d)=1/5$. The GP is $u_n=10(1/5)^{n-1}$. Even-indexed sum is $[10/5]/[1-1/25]=25/12$. Also $\ln|u_nu_{n+1}|=\ln100+(2n-1)\ln(1/5)$, an AP with difference $-2\ln5$.

Qualification: the question says infinite GP, while its solution assumes convergence. Without that assumption $d=0,a=10$ gives the constant GP, whose even-indexed series diverges and whose logarithmic sequence has difference 0.

An infinite GP need not have a finite sum. The printed solution supplies a convergence assumption absent from part (b) wording.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [2009/NYJC/Prelim/I/9] Question PDF pages 2; solution PDF pages 8, 9. The original files remain private QA records outside this native content.