Arithmetic Sequences and Series: past paper questions & worked solutions

The first term is $78$ and the common difference is $75-78=-3$.

$u_{18}=78+17(-3)=27$.

$S_{100}=\dfrac{100}{2}\{2(78)+99(-3)\}$.

$S_{100}=50(156-297)=50(-141)=-7050$.

The first term is $2$, the common difference is $3$, and the last term is $254$.

Use $254=2+(n-1)3$. Then $252=3(n-1)$, so $n-1=84$ and $n=85$.

$S_n=\dfrac n2(a+l)$, where $a$ is the first term and $l$ is the last term.

$S=\dfrac{85}{2}(2+254)=85\times128=10880$.

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.

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 $N=28$.

For an arithmetic series, $u_6=a+5d=2$.

$S_{10}=\dfrac{10}{2}(2a+9d)=-80$, so $2a+9d=-16$.

From $a+5d=2$, $a=2-5d$. Substitute into $2a+9d=-16$.

$2(2-5d)+9d=-16$, so $4-d=-16$ and $d=20$. Then $a=2-5(20)=-98$.

$S_n=\dfrac n2(2a+(n-1)d)=\dfrac n2(-196+20n-20)=10n^2-108n$.

We need $10n^2-108n>8000$.

$10n^2-108n-8000=0$ gives $5n^2-54n-4000=0$.

$n=\dfrac{54+\sqrt{82916}}{10}=34.195\ldots$.

Therefore the smallest integer value with $S_n>8000$ is $n=35$.

The first term is $a$ and the common difference is $4$.

$S_{10}=\dfrac{10}{2}\left(2a+9\cdot4\right)=360$.

$5(2a+36)=360$, so $10a+180=360$ and $a=18$.

$S_n=\dfrac n2\left(2(18)+(n-1)4\right)=\dfrac n2(4n+32)=2n^2+16n$.

Since the total is $2146$, $2n^2+16n=2146$.

Dividing by $2$ and rearranging gives $n^2+8n-1073=0$.

$n=\dfrac{-8\pm\sqrt{8^2+4(1073)}}{2}=\dfrac{-8\pm66}{2}$, so $n=29$ or $n=-37$.

The number of rows must be positive, so there are $29$ rows.

The largest row is the last row: $18+(29-1)4=18+112=130$ seats.

$\left(2+\dfrac{kx}{8}\right)^7$ begins with $2^7+7(2^6)\left(\dfrac{kx}{8}\right)+\binom72(2^5)\left(\dfrac{kx}{8}\right)^2+\binom73(2^4)\left(\dfrac{kx}{8}\right)^3$.

This gives $128+56kx+\dfrac{21}{2}k^2x^2+\dfrac{35}{32}k^3x^3$.

The coefficients of $x$, $x^2$ and $x^3$ are $56k$, $\dfrac{21}{2}k^2$ and $\dfrac{35}{32}k^3$.

For three terms in an arithmetic progression, twice the middle term equals the sum of the outer terms: $21k^2=56k+\dfrac{35}{32}k^3$.

Since $k\ne0$, divide by $k$ and multiply by $32$: $672k=1792+35k^2$.

So $35k^2-672k+1792=0$. The discriminant is $672^2-4(35)(1792)=200704=448^2$.

$k=\dfrac{672\pm448}{70}$, giving $k=16$ or $k=\dfrac{16}{5}$.

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.

$S_n=a+(a+d)+(a+2d)+\cdots+(a+(n-1)d)$.

$S_n=(a+(n-1)d)+(a+(n-2)d)+\cdots+a$.

Each pair adds to $2a+(n-1)d$, and there are $n$ pairs. Hence $2S_n=n\{2a+(n-1)d\}$.

Dividing by $2$ gives $S_n=\dfrac n2\{2a+(n-1)d\}$.

$u_5=5(5)+3(-1)^5=25-3=22$.

$\sum_{n=1}^{59}u_n=5\sum_{n=1}^{59}n+3\sum_{n=1}^{59}(-1)^n$.

$5\sum_{n=1}^{59}n=5\cdot\dfrac{59\cdot60}{2}=8850$, and $\sum_{n=1}^{59}(-1)^n=-1$. Therefore the total is $8850-3=8847$.

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 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.

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 $5$.

Let $u=2\theta-10^\circ$. Then $\sin u=\dfrac13$.

Since $0\leq\theta<180^\circ$, $-10^\circ\leq u<350^\circ$. In this interval, $u=19.471\ldots^\circ$ or $u=160.529\ldots^\circ$.

$\theta=\dfrac{u+10^\circ}{2}$, giving $\theta=14.7^\circ$ and $85.3^\circ$ to one decimal place.

For three consecutive terms of an arithmetic sequence, second minus first equals third minus second. Hence $\dfrac1{\tan\alpha}-\sin\alpha=2\sin\alpha-\dfrac1{\tan\alpha}$.

This gives $\dfrac2{\tan\alpha}=3\sin\alpha$. Since $\dfrac1{\tan\alpha}=\dfrac{\cos\alpha}{\sin\alpha}$, we get $\dfrac{2\cos\alpha}{\sin\alpha}=3\sin\alpha$, so $2\cos\alpha=3\sin^2\alpha$.

Using $\sin^2\alpha=1-\cos^2\alpha$, $2\cos\alpha=3(1-\cos^2\alpha)$, so $3\cos^2\alpha+2\cos\alpha-3=0$.

$\cos\alpha=\dfrac{-1+\sqrt{10}}{3}$; the other root is less than $-1$. Since $\pi<\alpha<2\pi$ and cosine is positive, $\alpha$ is in the fourth quadrant.

$\alpha=2\pi-\cos^{-1}\left(\dfrac{-1+\sqrt{10}}{3}\right)=5.517$ radians to 3 decimal places.

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