Link an AP with a convergent GP · source reference

Printed question

An arithmetic series has first term $a$ and common difference $d$, where $a>0$ and $d$ is non-zero. The first, eighth and thirteenth terms of the arithmetic series are equal to the first three terms of a geometric series respectively.

(a) Show that $a=-\frac{49}2d$. [2]

(b) Show that the geometric series is convergent and find its sum to infinity, $S$, in terms of $a$. [3]

(c) Given that the sum of the first $n$ terms of the arithmetic series is 25 times the third term of the geometric series, find $n$. [3]

(d) Find the least $k$ such that the sum of the first $k$ terms of the geometric series is more than 98% of $S$. [3]

Printed working

(a) Equal GP ratios give

$$\frac{a+7d}a=\frac{a+12d}{a+7d}\Rightarrow a^2+14ad+49d^2=a^2+12ad\Rightarrow49d^2+2ad=0.$$

Since $d\ne0$, $49d+2a=0$, so $a=-\frac{49}2d$.

(b)

$$r=\frac{a+7d}a=\frac{-49d/2+7d}{-49d/2}=\frac57.$$

Since $|r|=\frac57<1$, the geometric series converges; $S=\frac a{1-5/7}=\frac72a$.

(c) $a=-\frac{49}2d\Rightarrow d=-\frac2{49}a$.

$$\frac n2\left[2a+(n-1)(-2a/49)\right]=25\left[a(5/7)^2\right].$$

Since $a\ne0$,

$$\frac n2[2+(n-1)(-2/49)]=\frac{625}{49}\Rightarrow n-\frac{n^2}{49}+\frac n{49}=\frac{625}{49}\Rightarrow n^2-50n+625=0\Rightarrow(n-25)^2=0.$$

Thus $n=25$.

Alternative:

$$\frac n2[2a+(n-1)d]=25(a+12d),$$ $$\frac n2\left[2(-49d/2)+(n-1)d\right]=25(-49d/2+12d)\Rightarrow\frac n2[-50d+nd]=25(-25d/2),$$ $$n^2d-50nd=-625d.$$

Since $d\ne0$, $n^2-50n+625=0$, so $n=25$.

(d) Let the first $k$ terms sum to $S_k$. Then

$$S_k>\frac{98}{100}S\Rightarrow\frac{a[1-(5/7)^k]}{1-5/7}>\frac{98}{100}(7a/2).$$

Since $a>0$, $1-(5/7)^k>98/100$, so $(5/7)^k<2/100$.

$$k\ln(5/7)<\ln(2/100)\Rightarrow k>\frac{\ln(2/100)}{\ln(5/7)}=11.627.$$

Least $k=12$.

Teacher notes

GP terms: $a,a+7d,a+12d$; common ratio $\frac{a+7d}a=\frac{a+12d}{a+7d}$.

Source tables and calculator displays

Alternative GC table:

$k$	$1-(5/7)^k$
11	0.97531 (<0.98)
12	0.98236 (>0.98)

Least $k=12$.

Printed answer key

(b) $\frac72a$; (c) 25; (d) least $k=12$

Mathematical review and qualifications

(a) Expansion gives $2ad+49d^2=0$. Since $d \ne0$, $a=-49d/2$.

(b) $d=-2a/49$ gives GP ratio $r=(a+7d)/a=5/7$. Since $|r|<1$, $S=a/(1-r)=7a/2$.

(c) The GP third term is $25a/49$. Hence $\frac n2[2a+(n-1)(-2a/49)]=625a/49$, or $n(50-n)=625$, so $(n-25)^2=0$ and $n=25$.

(d) $1-(5/7)^k>0.98$ gives $(5/7)^k<0.02$, so $k>11.626\ldots$ and least $k=12$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [2022/YIJC/Promo/8] Question PDF pages 1; solution PDF pages 2, 3, 4. The original files remain private QA records outside this native content.