Geometric Sequences and Series: past paper questions & worked solutions

For first term $a$ and common ratio $r$, $S_\infty=\dfrac{a}{1-r}=75$ and $S_3=\dfrac{a(1-r^3)}{1-r}=70.2$.

$\dfrac{S_3}{S_\infty}=1-r^3$, so $\dfrac{70.2}{75}=1-r^3$.

$0.936=1-r^3$, so $r^3=0.064$ and $r=0.4$.

Use $\dfrac{a}{1-0.4}=75$. Hence $a=75(0.6)=45$.

Jem's payments are arithmetic with first term $20$ and common difference $0.50$.

The month $100$ payment is $20+99(0.50)=69.50$, so Jem pays £69.50.

$S_{300}=\dfrac{300}{2}\{2(20)+299(0.50)\}=150(189.5)=28425$.

Let the common ratio be $r$. Since the first payment is $20$ and the month $300$ payment is $250$, $20r^{299}=250$.

Thus $r^{299}=12.5$ and $r=12.5^{1/299}$.

$S_B=20\dfrac{r^{300}-1}{r-1}$ with $r=12.5^{1/299}$, giving $S_B\approx27362.95$.

The difference is $28425-27362.95\approx1062.05$, which is £1,060 to the nearest £10.

An increase of $2\%$ each year gives the geometric ratio $r=1.02$.

From the start of 2019 to the start of 2032 there are $13$ yearly increases, so the population is $25000(1.02)^{13}$.

$25000(1.02)^{13}\approx32340.7$, so the model gives about $32341$ adults.

From 2019 to 2032 inclusive there are $14$ yearly populations, so the total number of adult payments is $25000\dfrac{1.02^{14}-1}{1.02-1}$.

The total money is $5\times25000\dfrac{1.02^{14}-1}{0.02}\approx1\,996\,766$, so this is about £1,996,766.

To the nearest £1000, this is £1,997,000.

For a geometric sequence, the third term is $ar^2=20$ and the fifth term is $ar^4=12.8$.

$\dfrac{ar^4}{ar^2}=\dfrac{12.8}{20}$, so $r^2=0.64$.

Since $r>0$, $r=0.8$.

Using $ar^2=20$, $a(0.8)^2=20$, so $a=\dfrac{20}{0.64}=31.25$.

$S_n=\dfrac{a(1-r^n)}{1-r}=\dfrac{31.25(1-0.8^n)}{0.2}=156.25(1-0.8^n)$.

We need $156.25(1-0.8^n)>156$.

$1-0.8^n>\dfrac{156}{156.25}=\dfrac{624}{625}$, so $0.8^n<\dfrac{1}{625}$.

$n\log(0.8)<\log\left(\dfrac{1}{625}\right)$. Since $\log(0.8)<0$, dividing reverses the inequality: $n>\dfrac{\log(1/625)}{\log(0.8)}=28.85\ldots$.

The smallest possible integer value is therefore $n=29$.

For a geometric sequence, $S_\infty=\dfrac{a}{1-r}$. The question gives $S_\infty=3a$.

$\dfrac{a}{1-r}=3a$. Since $u_2-u_4=16$, $a\ne0$, so $\dfrac{1}{1-r}=3$ and $r=\dfrac23$.

$u_2-u_4=ar-ar^3=ar(1-r^2)=16$.

$a\left(\dfrac23\right)\left(1-\dfrac49\right)=16$.

$a\cdot\dfrac23\cdot\dfrac59=16$, so $\dfrac{10a}{27}=16$ and $a=\dfrac{216}{5}=43.2$.

$S_{10}=\dfrac{a(1-r^{10})}{1-r}=3a\left(1-\left(\dfrac23\right)^{10}\right)$.

$S_{10}=3\left(\dfrac{216}{5}\right)\left(1-\left(\dfrac23\right)^{10}\right)=\dfrac{92840}{729}=127.352\ldots$, so $S_{10}=127$ to 3 significant figures.

$S_n=a+ar+ar^2+\cdots+ar^{n-1}$.

$rS_n=ar+ar^2+ar^3+\cdots+ar^n$.

$S_n-rS_n=a-ar^n=a(1-r^n)$.

$(1-r)S_n=a(1-r^n)$, so $S_n=\dfrac{a(1-r^n)}{1-r}$.

An $8\%$ decrease each year means the multiplier is $0.92$.

After $6$ years, the amount is $150(0.92)^6=90.953\ldots$, which is approximately $91$ litres.

At the end of $40$ years, the amounts are $150(0.92),150(0.92)^2,\ldots,150(0.92)^{40}$.

The total is $150(0.92)\dfrac{1-0.92^{40}}{1-0.92}=1663.581\ldots$.

To the nearest litre, the total amount is $1664$ litres.

The dormice model is $D=2000(1.03)^t$.

$D=2000(1.03)^6=2388.10\ldots$, so there are $2390$ dormice to 3 significant figures.

Let the vole model be $N=ab^t$.

At $t=4$, $ab^4=3690$. At $t=7$, $ab^7=3470$.

Dividing gives $b^3=\dfrac{3470}{3690}$, so $b=0.9797\ldots$, which is $0.98$ to 2 significant figures.

$a=\dfrac{3690}{b^4}=4005.18\ldots$, which is $4000$ to 2 significant figures.

So the model is approximately $N=4000(0.98)^t$.

Use $2000(1.03)^T=a b^T$ with the unrounded constants from the vole model.

$\left(\dfrac{1.03}{b}\right)^T=\dfrac{a}{2000}$, so $T=\dfrac{\log(a/2000)}{\log(1.03/b)}$.

Substituting $a=4005.18\ldots$ and $b=0.9797\ldots$ gives $T=13.875\ldots$, so $T=13.9$ to one decimal place.

$\displaystyle \sum_{r=1}^{\infty}6(0.25)^r$ has first term $6(0.25)=1.5$ and common ratio $0.25$.

$S_\infty=\dfrac{1.5}{1-0.25}=\dfrac{1.5}{0.75}=2$.

$u_1=3$. Then $u_2=\dfrac{3-3}{3-2}=0$.

$u_3=\dfrac{0-3}{0-2}=\dfrac32$, and $u_4=\dfrac{\frac32-3}{\frac32-2}=3$.

Since $u_4=3=u_1$, the terms repeat as $3,0,\dfrac32,3,0,\dfrac32,\ldots$.

The repeating cycle has length $3$, so the order is $3$.

One full cycle sums to $3+0+\dfrac32=\dfrac92$.

Since $70=23\cdot3+1$, the first $70$ terms contain $23$ complete cycles and one extra term $u_1=3$.

$\displaystyle \sum_{n=1}^{70}u_n=23\left(\dfrac92\right)+3=\dfrac{207}{2}+\dfrac{6}{2}=\dfrac{213}{2}$.

$u_2=u_1r$, $u_3=u_1r^2$, and $u_4=u_1r^3$.

Since $u_4=8$, $u_1r^3=8$, so $u_1r=\dfrac{8}{r^2}$.

$u_2+u_3=u_1r(1+r)=6$.

Substitute $u_1r=\dfrac8{r^2}$ to get $\dfrac8{r^2}(1+r)=6$. Hence $8+8r=6r^2$, so $3r^2-4r-4=0$.

$3r^2-4r-4=0$ gives $r=2$ or $r=-\dfrac23$.

A sum to infinity requires $|r|<1$, so $r=-\dfrac23$.

$u_1r^3=8$, so $u_1\left(-\dfrac23\right)^3=8$ and $u_1=-27$.

$S_\infty=\dfrac{u_1}{1-r}=\dfrac{-27}{1+2/3}=-\dfrac{81}{5}$.

Let the yearly ratio be $r$. After 3 years, $30000r^3=34000$.

$r^3=\dfrac{17}{15}$, so $r=\left(\dfrac{17}{15}\right)^{1/3}\approx1.04257$. Therefore $p=(r-1)\times100\approx4.26$.

After $N$ years the model gives $30000r^N$. We need $30000r^N>75000$, so $r^N>2.5$.

$N\log r>\log(2.5)$, so $N>\dfrac{\log(2.5)}{\log r}$.

Using $r\approx1.04257$, $\dfrac{\log(2.5)}{\log r}\approx21.98$. Hence the smallest integer value is $N=22$.

The first term is $10$ and the common ratio is $\dfrac{8}{10}=0.8$.

$S_\infty=\dfrac{10}{1-0.8}=50$.

The $k$th term is $10(0.8)^{k-1}$.

$10(0.8)^{k-1}<0.0005$, so $(0.8)^{k-1}<0.00005$.

Taking logarithms gives $(k-1)\log(0.8)<\log(0.00005)$. Since $\log(0.8)<0$, $k-1>\dfrac{\log(0.00005)}{\log(0.8)}=44.381\ldots$.

Thus $k>45.381\ldots$, so the smallest possible value is $k=46$.

For $850+843+836+\cdots$, the first term is $850$ and the common difference is $-7$.

$S_n=\dfrac n2\{2(850)+(n-1)(-7)\}=\dfrac n2(1707-7n)$.

This quadratic has its maximum near $n=\dfrac{1707}{14}=121.928\ldots$, so check $n=122$.

$S_{122}=\dfrac{122}{2}(1707-854)=61(853)=52033$.

Model A is arithmetic. From year 1 to year 12 there are $11$ equal increases, so $d=\dfrac{4000-300}{11}=\dfrac{3700}{11}$.

Year 4 is $300+3d=300+3\left(\dfrac{3700}{11}\right)=1309.09\ldots$, so the amount is $1310$ tonnes to the nearest $10$ tonnes.

Model B is geometric. If the ratio is $r$, then $300r^{11}=4000$.

$r=\left(\dfrac{4000}{300}\right)^{1/11}=1.2655\ldots$.

Year 2 is $300r=379.65\ldots$, so the amount is $380$ tonnes to the nearest $10$ tonnes.

$S_A=\dfrac{12}{2}(300+4000)=25800$ tonnes.

$S_B=300\dfrac{r^{12}-1}{r-1}=17935.20\ldots$ tonnes.

The difference is $25800-17935.20\ldots=7864.79\ldots$, which is $7860$ tonnes to the nearest $10$ tonnes.

For three consecutive terms of a geometric sequence, the middle term squared equals the product of the outer terms.

So $k^2=(2k-15)(k+4)$.

Expanding gives $k^2=2k^2-7k-60$, hence $k^2-7k-60=0$.

$k^2-7k-60=(k-12)(k+5)=0$, so $k=12$ or $k=-5$. The download numbers must be positive, so $k=12$.

The first three monthly values are $9$, $12$ and $16$ thousand. The common ratio is $\dfrac{12}{9}=\dfrac43$.

The 4th month is $16\cdot\dfrac43=\dfrac{64}{3}$ thousand downloads, which is about $21300$ downloads.

In thousands, the total after $N$ months is $S_N=\dfrac{9\left(\left(\dfrac43\right)^N-1\right)}{\dfrac43-1}=27\left(\left(\dfrac43\right)^N-1\right)$.

Three million downloads is $3000$ thousand, so we need $27\left(\left(\dfrac43\right)^N-1\right)>3000$.

$\left(\dfrac43\right)^N>\dfrac{1009}{9}$. Taking logarithms gives $N>\dfrac{\log(1009/9)}{\log(4/3)}=16.405\ldots$.

The first integer month for which the total exceeds $3$ million is $N=17$.

For the first $20$ hits, the additional distances form an arithmetic sequence with first term $100$ and common difference $-2$.

The twentieth additional distance is $100+19(-2)=62$ mm.

$S_{20}=\dfrac{20}{2}(100+62)=1620$ mm.

After the twentieth hit, the distances follow the same ratio, so the twenty second distance is $62r^2$.

$62r^2=60$, hence $r=\sqrt{\dfrac{30}{31}}=0.984$ to 3 decimal places.

More than $3$ m means more than $3000$ mm. After $20$ hits, the remaining amount to exceed is $3000-1620=1380$ mm.

If there are $m$ further hits, their total is $62r+62r^2+\cdots+62r^m=62r\dfrac{1-r^m}{1-r}$.

Using $r=\sqrt{30/31}$, $62r\dfrac{1-r^{27}}{1-r}$ is not enough, but $62r\dfrac{1-r^{28}}{1-r}$ makes the total exceed $3000$ mm.

Therefore $m=28$ further hits are needed after the first $20$, so $N=20+28=48$.

For Model A, let the common difference be $d$. Since year 15 is the 15th term, $7500=12400+14d$.

$d=\dfrac{7500-12400}{14}=-350$. Year 5 is $12400+4(-350)=11000$ tonnes.

$S_{15}=\dfrac{15}{2}(12400+7500)=149250$ tonnes.

Let the common ratio be $r$. Then $7500=12400r^{14}$, so $r=\left(\dfrac{7500}{12400}\right)^{1/14}\approx0.964723$.

Year 10 is $12400r^9\approx8975.26$, which is $8980$ tonnes to the nearest 10 tonnes.

Since $0<r<1$, the sum to infinity is $S_\infty=\dfrac{12400}{1-r}\approx351508.1$, so the limit is $351500$ tonnes to the nearest 100 tonnes.

The nth term of the first sequence is $4\cdot6^{n-1}$.

We need $4\cdot6^{n-1}>10^{100}$, so $n-1>\dfrac{100-\log_{10}4}{\log_{10}6}=127.736\ldots$.

Thus $n$ must be greater than $128.736\ldots$, so the minimum possible value is $129$.

For the second sequence, $ar=-6$, so $a=-\dfrac6r$.

$\dfrac{a}{1-r}=25$. Substituting $a=-\dfrac6r$ gives $-6=25r(1-r)$.

Rearranging gives $25r^2-25r-6=0$.

$25r^2-25r-6=0$ gives $r=\dfrac65$ or $r=-\dfrac15$.

A sum to infinity exists only when $|r|<1$, so $r=-\dfrac15$.

Since $ar=-6$, $a=30$.

$S_4=30\dfrac{1-(-1/5)^4}{1-(-1/5)}=\dfrac{624}{25}$.

Plan A has $12$ Sunday runs from $15$ km to $37$ km, so there are $11$ equal increases. The common difference is $\dfrac{37-15}{11}=2$.

Week $5$ is the fifth term, so the distance is $15+4(2)=23\text{ km}$.

For plan B, $15r^{11}=37$, so $r=\left(\dfrac{37}{15}\right)^{1/11}$.

The fifth Sunday distance is $15r^4=15\left(\dfrac{37}{15}\right)^{4/11}=20.829\ldots$, so it is $20.8\text{ km}$.

The arithmetic total is $\dfrac{12}{2}(15+37)=312$ km.

The total distance condition gives $312+12x\leq360$, so $x\leq4$. The maximum integer is $4$.

The geometric total is $15\dfrac{r^{12}-1}{r-1}=294.185\ldots$ km.

Now $294.185\ldots+12x\leq360$, so $x\leq5.484\ldots$. The maximum integer is $5$.

The 2nd term of $(3+2x)^6$ is $\binom61 3^5(2x)=6\cdot243\cdot2x=2916x$.

The 3rd term is $\binom62 3^4(2x)^2=15\cdot81\cdot4x^2=4860x^2$.

The 5th term is $\binom64 3^2(2x)^4=15\cdot9\cdot16x^4=2160x^4$.

For three consecutive terms in a geometric series, the middle term squared equals the product of the two outer terms.

$(4860x^2)^2=(2916x)(2160x^4)$.

For the non-zero geometric case, divide by $x^4$: $4860^2=2916\cdot2160\,x$, so $x=\dfrac{4860^2}{2916\cdot2160}=\dfrac{15}{4}$.

In the different geometric series, first term $=\sin^2\theta$, common ratio $=2\cos\theta$, and $S_\infty=\dfrac85$.

$\dfrac{\sin^2\theta}{1-2\cos\theta}=\dfrac85$.

$5\sin^2\theta=8(1-2\cos\theta)$. With $\sin^2\theta=1-\cos^2\theta$, this becomes $5(1-\cos^2\theta)=8-16\cos\theta$.

Rearranging gives $5\cos^2\theta-16\cos\theta+3=0$.

$5\cos^2\theta-16\cos\theta+3=0$ gives $\cos\theta=3$ or $\cos\theta=\dfrac15$. The possible value is $\cos\theta=\dfrac15$.

The 2nd term is first term times common ratio: $\sin^2\theta(2\cos\theta)=\left(1-\dfrac{1}{25}\right)\cdot\dfrac25=\dfrac{24}{25}\cdot\dfrac25=\dfrac{48}{125}$.

The first term is $a$ and the $n$th term is $a+(n-1)d$.

$S_n=\dfrac n2(\text{first term}+\text{last term})$.

Therefore $S_n=\dfrac n2\{a+[a+(n-1)d]\}=\dfrac n2\{2a+(n-1)d\}$.

For $900+892+884+\cdots+500$, the first term is $900$, the common difference is $-8$, and the last term is $500$.

$500=900+(n-1)(-8)$, so $n-1=50$ and $n=51$.

$S=\dfrac{51}{2}(900+500)=35700$.

For $k+4$, $k-2$, $11-k$ to be consecutive geometric terms, $(k-2)^2=(k+4)(11-k)$.

$k^2-4k+4=-k^2+7k+44$, so $2k^2-11k-40=0$.

$2k^2-11k-40=0$ gives $k=8$ or $k=-\dfrac52$.

If $k=8$, the terms are $12,6,3$ and the common ratio is $\dfrac12$, so the series is convergent. If $k=-\dfrac52$, the common ratio is $-3$, so it is not convergent.

Use $k=8$: $S_\infty=\dfrac{12}{1-1/2}=24$.