A half-tail identity and triangular rows · source reference

Printed question

(a) The first 2 terms of a GP are $p,q$, where $q<p,p\ne0$. If the sum of the first $n$ terms equals the sum to infinity of the remaining terms, prove $p^n=2q^n$. [3]

(b) The AP $2,4,6,8,10,12,\ldots$ is arranged in rows: 1st row: 2; 2nd: 4,6; 3rd: 8,10,12; 4th: 14,16,18,20; … (native table below).

(i) Show the first term in row $n$ is $n^2-n+2$. [2]

(ii) Find the sum of all terms from row 1 to row $n-1$. [3]

Printed working

(a) Common ratio $r=q/p$. $S_n=S_{\infty}-S_n\Rightarrow2S_n=S_{\infty}$.

$$2\frac{p[1-(q/p)^n]}{1-q/p}=\frac p{1-q/p}\Rightarrow2[1-(q/p)^n]=1\Rightarrow2(q/p)^n=1\Rightarrow p^n=2q^n.$$

(b)(i) Total terms before row $n$:

$$1+2+\cdots+(n-1)=\frac{n-1}2[1+(n-1)]=\frac{n(n-1)}2.$$

First in row $n$ is term $n(n-1)/2+1$ of the AP, hence

$$2+[n(n-1)/2+1-1](2)=n^2-n+2.$$

Alternative: first terms of rows are $2,4,8,14,\ldots$; differences $2,4,6,\ldots$ form AP with $a=2,d=2$. Sum of first $n-1$ differences:

$$\frac{n-1}2[2(2)+(n-2)2]=n^2-n.$$

Add first term 2, giving $n^2-n+2$.

(b)(ii) With $N=n(n-1)/2$ total terms, first 2 and last $n^2-n+2-2$,

$$\frac N2[2+(n^2-n+2-2)]=\frac{n(n-1)(n^2-n+2)}4.$$

Teacher notes

GP formulas $S_n=a(1-r^n)/(1-r)$, $S_{\infty}=a/(1-r)$, here $a=p,r=q/p$.

AP sum $\frac n2(a+l)$, where $a$ is first, $l$ is last; for counting preceding rows $a=1,l=n-1,n-1$ terms. For even-number AP $a=2,d=2,N=n(n-1)/2+1$.

For the final sum, last term of row $n-1$ is the first term of row $n$ minus 2; total term count $N=n(n-1)/2$.

Source tables and calculator displays

Row	Terms
1st	2
2nd	4, 6
3rd	8, 10, 12
4th	14, 16, 18, 20
…	…

Printed answer key

(b)(ii) $\frac{n(n-1)(n^2-n+2)}4$

Mathematical review and qualifications

(a) The nonzero convergent GP has $r=q/p$. Tail equality gives $p(1-r^n)/(1-r)=pr^n/(1-r)$, so $r^n=1/2$, hence $p^n=2q^n$.

(b)(i) Let $N=n(n-1)/2$ be the number of terms before row $n$. Its first term is $2(N+1)=n^2-n+2$. (ii) The sum of the first $N$ even integers is $N(N+1)=n(n-1)(n^2-n+2)/4$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [2009/DHS/Prelim/I/9] Question PDF pages 3; solution PDF pages 14, 15. The original files remain private QA records outside this native content.