Modelling with Sequences and Series: past paper questions & worked solutions

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.

$D=8+5\sin\left(\dfrac{\pi(2)}{6}+3\right)=8+5\sin\left(\dfrac{\pi}{3}+3\right)$.

$D=4.064\ldots$, so the depth at 2 am is just over $4$ metres.

$6=8+5\sin\left(\dfrac{\pi t}{6}+3\right)$, so $\sin\left(\dfrac{\pi t}{6}+3\right)=-\dfrac25$.

After midday means $t>12$. Then $\dfrac{\pi t}{6}+3$ is greater than $2\pi+3$.

The first angle after $2\pi+3$ with sine $-\dfrac25$ is $3\pi+\sin^{-1}\left(\dfrac25\right)$.

$\dfrac{\pi t}{6}+3=3\pi+\sin^{-1}\left(\dfrac25\right)$, so $t=13.056\ldots$ hours after midnight.

$0.056\ldots$ hours is about $3.38$ minutes, so the time to the nearest minute is $13{:}03$, or $1{:}03$ pm.

The $x$ values are $0,3,6,9,12,15$, so the trapezium width is $h=3$.

$\text{Area}\approx \dfrac32\left[0+0+2(1.52+2.74+3.12+3.08)\right]$.

This gives $31.38\text{ m}^2$.

The speed is $1.5\text{ m s}^{-1}$, so in one minute the water travels $1.5\times60=90\text{ m}$.

Volume per minute $\approx31.38\times90=2824.2\text{ m}^3$.

To 2 significant figures, this is $2.8\times10^3\text{ m}^3$.

It is an underestimate, because the curve between the plotted points lies below the straight-line trapezium chords in the diagram, so the true cross-sectional area is larger.

The first term is $480$ and the common difference is $465-480=-15$.

$u_{14}=480+13(-15)=285$, so $285$ tonnes were extracted in year $14$.

The total after $N$ years is $S_N=\dfrac N2\{2(480)+(N-1)(-15)\}$.

$\dfrac N2(975-15N)=7770$.

Multiplying by $2$ and rearranging gives $15N^2-975N+15540=0$, so $N^2-65N+1036=0$.

$N=28$ or $37$. Since extraction amounts would become negative before year $37$, the stopping year is $N=28$.

$P(10)=0.95+2^{-2}+2^2-(10-12)^2=0.95+0.25+4-4=1.20$.

$P(11.5)=0.95+2^{-0.5}+2^{0.5}-(11.5-12)^2=2.821\ldots$.

The time values increase by $0.5$, so $h=0.5$.

The estimate is $\dfrac{0.5}{2}\{1.20+2.95+2(1.882+2.45+2.821\ldots)\}$.

This gives $4.614\ldots$, so the estimated energy is $4.61\text{ kWh}$ to 2 decimal places.

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.

The weekly amounts are $5$, $5.25$, $5.50$, and so on. The increase is a constant $0.25$, so an arithmetic series should be used.

The first amount is $5$ and the common difference is $0.25$, so the amount saved in week $n$ is $5+0.25(n-1)$.

After $n$ weeks, $S_n=\dfrac n2\{2(5)+(n-1)(0.25)\}$.

Adina needs $S_n\geq350$, so $\dfrac n2(9.75+0.25n)\geq350$.

This is $n^2+39n-2800\geq0$. The positive root is $36.89\ldots$, so the first whole number that works is $n=37$.

After $36$ weeks the total is below $350$, and after $37$ weeks it is $351.50$, so it takes $37$ weeks.

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

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

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

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

When $t=3$, $w=10$. So $10=\log_a(3+5)-\log_a4=\log_a8-\log_a4$.

$\log_a8-\log_a4=\log_a\left(\dfrac84\right)=\log_a2$, so $10=\log_a2$.

$10=\log_a2$ means $a^{10}=2$, hence $a=2^{1/10}=1.071773\ldots=1.072$ to 3 decimal places.

Using $a=1.072$, $w=\log_{1.072}(t+5)-\log_{1.072}4$.

$w=\log_{1.072}\left(\dfrac{t+5}{4}\right)$.

Changing from logarithmic form to exponential form gives $(1.072)^w=\dfrac{t+5}{4}$, so $t=4(1.072)^w-5$.

When $w=15$, $t=4(1.072)^{15}-5=6.3496\ldots$, so $t=6.35$ months to 3 significant figures.

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 $N=22$.

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 $k=1$, $N=\lambda^t$ with $0<\lambda<1$. The graph starts at $(0,1)$, decreases as $t$ increases, and tends towards $0$.

After $5700$ years, $N=\dfrac{k}{2}$, so $k\lambda^{5700}=\dfrac{k}{2}$.

$\lambda^{5700}=\dfrac12$, hence $\lambda=\left(\dfrac12\right)^{1/5700}=0.999878\ldots$, so $\lambda=0.999878$ to 6 decimal places.

For item A, $k=15$ and $t=3250$, so $N=15\lambda^{3250}$.

$N=15(0.9998784026\ldots)^{3250}=10.103\ldots$, so the current amount is $10.1$ g to 3 significant figures.

For item B, $18=25\lambda^t$, so $\lambda^t=\dfrac{18}{25}$.

$t\log\lambda=\log\left(\dfrac{18}{25}\right)$, so $t=\dfrac{\log(18/25)}{\log\lambda}$.

Using $\lambda=(1/2)^{1/5700}$ gives $t=2701\ldots$, so the age is $2700$ years to the nearest $100$ years.

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 volume is $(3x)(x)(y)=3x^2y$. Since the capacity is $120$, $3x^2y=120$, so $y=\dfrac{40}{x^2}$.

The open top means the metal area is the base plus four sides: $A=3x^2+2(3xy)+2(xy)$.

$A=3x^2+8xy=3x^2+8x\left(\dfrac{40}{x^2}\right)=3x^2+\dfrac{320}{x}$. Hence $P=3$ and $Q=320$.

$\dfrac{dA}{dx}=6x-320x^{-2}=6x-\dfrac{320}{x^2}$.

Set $\dfrac{dA}{dx}=0$: $6x=\dfrac{320}{x^2}$, so $6x^3=320$.

$x^3=\dfrac{160}{3}$, so $x=\sqrt[3]{\dfrac{160}{3}}=3.764\ldots$. To $3$ significant figures, $x=3.76$.

$\dfrac{d^2A}{dx^2}=6+640x^{-3}=6+\dfrac{640}{x^3}$.

For $x>0$, $6+\dfrac{640}{x^3}>0$, so the stationary value found in part (b) is a minimum.

Each end is a right-angled triangle with perpendicular sides $3x$ and $2x$, so its area is $\dfrac12(3x)(2x)=3x^2$.

The volume is cross-section area times length, so $3x^2L=12$ and $L=\dfrac4{x^2}$.

There are two triangular ends with total area $6x^2$, and two rectangular sides with total area $3xL+2xL=5xL$.

$S=6x^2+5x\left(\dfrac4{x^2}\right)=6x^2+\dfrac{20}{x}$. Hence $P=6$ and $Q=20$.

$\dfrac{dS}{dx}=12x-20x^{-2}=12x-\dfrac{20}{x^2}$.

Set $\dfrac{dS}{dx}=0$: $12x=\dfrac{20}{x^2}$, so $x^3=\dfrac53$.

At $x=\sqrt[3]{\dfrac53}$, $S=6x^2+\dfrac{20}{x}=25.3029\ldots$, so the minimum value is $25.3$ to one decimal place.

$\dfrac{d^2S}{dx^2}=12+40x^{-3}=12+\dfrac{40}{x^3}$.

Since $x>0$, $12+\dfrac{40}{x^3}>0$, so the stationary value found is a minimum.

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 $5$.

$5\sin x\tan x+13=\cos x$ becomes $\dfrac{5\sin^2x}{\cos x}+13=\cos x$.

$5\sin^2x+13\cos x=\cos^2x$.

$5(1-\cos^2x)+13\cos x=\cos^2x$, so $6\cos^2x-13\cos x-5=0$.

$(2\cos x-5)(3\cos x+1)=0$, so $\cos x=\dfrac52$ or $\cos x=-\dfrac13$.

$\cos x=\dfrac52$ is impossible, so $\cos x=-\dfrac13$. For $0<x\le\pi$, $x=1.9106\ldots$, so $x=1.91$ radians to 3 significant figures.

At $t=6$, $20=10+12\sin(6k+18)^\circ$, so $\sin(6k+18)^\circ=\dfrac56$.

$\sin^{-1}\left(\dfrac56\right)=56.442\ldots^\circ$. Since $0<k<20$, $18<6k+18<138$, so $6k+18=56.442\ldots$ or $123.557\ldots$.

This gives $k=6.407\ldots$ or $k=17.592\ldots$, so $k=6.41$ or $17.59$ to 2 decimal places.

Given further that $0<k<10$, use $k=6.407\ldots$.

The maximum value of $\sin$ is $1$, so the maximum temperature is $10+12=22^\circ\text{C}$.

The maximum occurs when $kt+18=90$, so $t=\dfrac{72}{6.407\ldots}=11.2375\ldots$ hours after midnight.

$0.2375\ldots$ hours is $14.25\ldots$ minutes, so the maximum occurs at about $11{:}14$ am.