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.

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

$u_3=3u_2-2u_1$.

$4=3u_2-2(7)$, so $4=3u_2-14$.

$3u_2=18$, hence $u_2=6$.

$u_4=3u_3-2u_2=3(4)-2(6)=0$.

$\displaystyle \sum_{r=1}^{4}(u_r+2r)=(7+2)+(6+4)+(4+6)+(0+8)$.

The sum is $9+10+10+8=37$.

$u_2=2-\dfrac4{u_1}=2-\dfrac43=\dfrac23$.

$u_3=2-\dfrac4{u_2}=2-\dfrac4{2/3}=2-6=-4$.

$u_4=2-\dfrac4{u_3}=2-\dfrac4{-4}=3$.

The sequence starts $3,\dfrac23,-4,3,\dfrac23,-4,\ldots$, so it repeats every 3 terms.

$3+\dfrac23-4=-\dfrac13$.

There are $33$ complete cycles in the first $99$ terms, then one extra term $u_{100}=3$.

$\displaystyle \sum_{r=1}^{100}u_r=33\left(-\dfrac13\right)+3=-11+3=-8$.

$u_2=ku_1+3=6k+3$.

$u_3=ku_2+3=k(6k+3)+3=6k^2+3k+3$.

$u_1+u_2+u_3=117$, so $6+(6k+3)+(6k^2+3k+3)=117$.

$6k^2+9k+12=117$, hence $2k^2+3k-35=0$.

$(2k-7)(k+5)=0$. Since $k$ is positive, $k=\dfrac72$.

$(1-4x)^7=1+7(-4x)+\binom72(-4x)^2+\binom73(-4x)^3+\cdots$.

The first four terms are $1-28x+336x^2-2240x^3$.

In $(5+kx)(1-4x)^7$, the $x^2$ coefficient comes from $5(336x^2)$ and $(kx)(-28x)$.

$5(336)-28k=1316$, so $1680-28k=1316$.

$28k=364$, hence $k=13$.

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 $a=75(0.6)=45$.

Since $2=\log_3 9$, the equation becomes $\log_3(4x)+\log_3 9=\log_3(5x+7)$.

$\log_3(4x)+\log_3 9=\log_3(36x)$.

Therefore $36x=5x+7$, so $31x=7$ and $x=\dfrac{7}{31}$.

Let $L=\log_a y$. Then $\log_a(y^r)=r\log_a y=rL$.

$\displaystyle \sum_{r=1}^{2}\log_a(y^r)=L+2L=3L$.

$\displaystyle \sum_{r=1}^{2}(\log_a y)^r=L+L^2$.

So $3L=L+L^2$, which gives $L^2-2L=0$ and $L(L-2)=0$.

Since $a>1$ and $y>1$, $\log_a y>0$, so $L\ne0$. Hence $L=2$ and $y=a^2$.

$u_1=7$, $u_2=(-1)^1u_1+k=k-7$, $u_3=(-1)^2u_2+k=2k-7$, and $u_4=(-1)^3u_3+k=7-k$.

$u_5=(-1)^4u_4+k=(7-k)+k=7$.

$\displaystyle \sum_{r=1}^{4}u_r=7+(k-7)+(2k-7)+(7-k)=2k$.

Since $\displaystyle \sum_{r=1}^{4}u_r=30$, we have $2k=30$, so $k=15$.

With $k=15$, the first four terms are $7,8,23,-8$, and their sum is $30$. Since $u_5=u_1$, the pattern repeats every four terms.

There are $37$ complete blocks in the first $150$ terms, with the first two terms left over. Hence $\displaystyle \sum_{r=1}^{150}u_r=37(30)+7+8=1125$.

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.

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 $n=29$.

$a_1=\cos^2\left(\dfrac{\pi}{3}\right)=\left(\dfrac12\right)^2=\dfrac14$.

$a_2=\cos^2\left(\dfrac{2\pi}{3}\right)=\left(-\dfrac12\right)^2=\dfrac14$.

$a_3=\cos^2(\pi)=(-1)^2=1$.

The values of $\cos^2\left(\dfrac{n\pi}{3}\right)$ repeat every 3 terms as $\dfrac14,\dfrac14,1$.

$\sum_{n=1}^{50} n=\dfrac{50\cdot51}{2}=1275$.

There are $16$ complete groups of 3 terms in the first $48$ terms, then the first 2 terms repeat. Hence the cosine-square sum is $16\left(\dfrac14+\dfrac14+1\right)+\dfrac14+\dfrac14=24+\dfrac12=\dfrac{49}{2}$.

$\sum_{n=1}^{50}\left(n+\cos^2\left(\dfrac{n\pi}{3}\right)\right)=1275+\dfrac{49}{2}=\dfrac{2599}{2}$.

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.

Since $a_1=p-3$, we have $a_1+3=p$. Therefore $a_2=2(a_1+3)^2-7=2p^2-7$.

$a_2+3=2p^2-4$, so $a_3=2(2p^2-4)^2-7=8(p^2-2)^2-7$.

$a_1+a_2+a_3=p+15$.

$(p-3)+(2p^2-7)+8(p^2-2)^2-7=p+15$.

Cancelling $p$ and expanding gives $8p^4-30p^2=0$, so $2p^2(4p^2-15)=0$.

If $p=0$, then $a_2=-7$. If $p^2=\dfrac{15}{4}$, then $a_2=2\cdot\dfrac{15}{4}-7=\dfrac12$.

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

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

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 $8850-3=8847$.

Since $\log_a b=k$, we can replace every $\log_a b$ by $k$.

$\log_a\left(\dfrac{\sqrt a}{b}\right)=\log_a(a^{1/2})-\log_a b=\dfrac12-k$.

$\log_a(a^2b)=\log_a a^2+\log_a b=2+k$, and $\log_a(b^3)=3\log_a b=3k$.

Therefore $\dfrac{\log_a(a^2b)}{\log_a(b^3)}=\dfrac{k+2}{3k}$.

$\log_a(b^n)=n\log_a b=nk$, so $k+\log_a(b^n)=k+nk=(n+1)k$.

$\sum_{n=1}^{50}(n+1)k=k\left(\sum_{n=1}^{50}n+50\right)=k(1275+50)=1325k$.

$\displaystyle \sum_{r=1}^{\infty}6(0.25)^r$ 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 $u_1=3$.

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

Using $u_{n+1}=b-au_n$ and $u_1=3$, we get $u_2=b-a(3)=b-3a$.

$u_3=b-au_2=b-a(b-3a)=b-ab+3a^2$.

$u_1+u_2+u_3=153$, so $3+(b-3a)+(b-ab+3a^2)=153$.

Using $b=a+9$, the left side becomes $3+(a+9)-3a+(a+9)-a(a+9)+3a^2$.

This is $2a^2-10a+21$, so $2a^2-10a+21=153$.

$2a^2-10a-132=0$. Dividing by $2$ gives $a^2-5a-66=0$.

$a^2-5a-66=(a-11)(a+6)=0$, so $a=11$ or $a=-6$.

If $a=11$, then $b=20$ and $u_2=20-3(11)=-13$. If $a=-6$, then $b=3$ and $u_2=3-3(-6)=21$.

The larger possible value of $u_2$ is $21$.

$u_2=u_1r$, $u_3=u_1r^2$, 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 $u_1=-27$.

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

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 $n=122$.

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

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

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