Chapter 8: Predicting What Comes Next: Exploring Sequences and Progressions (Sequences and Progressions)

Each term is 3 more than the one before, so the next four terms are 16, 19, 22, 25.

Running totals (each new total = previous total + next term):

Next four terms: 16, 19, 22, 25. Sums: 1, 5, 12, 22, 35, 51, 70, 92, 117, 145.

The triangular numbers are 1, 3, 6, 10, ...; each is the previous one plus the next natural number ($t_n=t_{n-1}+n$).

$$t_5=10+5=15,t_6=15+6=21,t_7=21+7=28,t_8=28+8=36$$

t₅ = 15, t₆ = 21, t₇ = 28, t₈ = 36

$$u_{53}=106-1=105,u_{108}=216-1=215,u_{1170}=2340-1=2339$$

105, 215 and 2339

(i) $t_1=-4$, $t_2=-1$, $t_3=2$, $t_{12}=29$, $t_{18}=47$, $t_{50}=143$.

(ii) $3n-7=332\Rightarrow 3n=339\Rightarrow n=113$.

(iii) $3n-7=557\Rightarrow 3n=564\Rightarrow n=188$, a natural number. So yes, 557 is the 188th term.

(i) −4, −1, 2, 29, 47, 143 (ii) the 113th term (iii) Yes, 557 is the 188th term.

Substitute n = 1, 2, 3, 4, 5:

• (i) −1, 2, 5, 8, 11
• (ii) −3, −8, −13, −18, −23
• (iii) $1-2+3=2$; $4-4+3=3$; $9-6+3=6$; $16-8+3=11$; $25-10+3=18$: so 2, 3, 6, 11, 18

(i) −1, 2, 5, 8, 11 (ii) −3, −8, −13, −18, −23 (iii) 2, 3, 6, 11, 18

$$t_{10}=50-3=47,t_{15}=75-3=72$$

t₁₀ = 47 and t₁₅ = 72

$$5n-3=97\Rightarrow 5n=100\Rightarrow n=205n-3=172\Rightarrow 5n=175\Rightarrow n=35$$

Both values of n are natural numbers.

Yes, both: 97 is the 20th term and 172 is the 35th term.

$$5n-3=607\Rightarrow 5n=610\Rightarrow n=122$$

607 is the 122nd term.

First five terms: −5, −2, 1, 4, 7. It is an AP with a = −5, d = 3, so $t_n=-5+3(n-1)=3n-8$.

$$3n-8=52\Rightarrow 3n=60\Rightarrow n=20$$

−5, −2, 1, 4, 7; yes, 52 is the 20th term.

Each term is the sum of the three before it:

$$T_4=4+2+1=7,T_5=7+4+2=13,T_6=13+7+4=24,T_7=24+13+7=44,T_8=44+24+13=81$$

T₄ = 7, T₅ = 13, T₆ = 24, T₇ = 44, T₈ = 81

(i) Differences: 5 − 2 = 8 − 5 = 11 − 8 = 3, constant, so it is an AP with a = 2, d = 3: $t_n=2+3(n-1)=3n-1$.

(ii) Differences: −1 − (−5) = 3 − (−1) = 7 − 3 = 4, constant: an AP with a = −5, d = 4: $t_n=-5+4(n-1)=4n-9$.

When we plot the points (n, tₙ), they lie on a straight line, just as for Fig. 8.4. The slope of the line is the common difference d (3 and 4 here), because the nth term $a+(n-1)d$ is a linear expression in n.

(i) AP with d = 3, tₙ = 3n − 1 (ii) AP with d = 4, tₙ = 4n − 9. The points lie on straight lines.

(i) $a=\frac{1}{2}$, $d=\frac{5}{2}-\frac{1}{2}=2$:

$$t_n=\frac{1}{2}+2(n-1)=2n-\frac{3}{2}=\frac{4n-3}{2}$$

(ii) $a=1.5$, $d=2$: $t_n=1.5+2(n-1)=2n-0.5$.

(i) $t_n=\frac{4n-3}{2}$ (ii) $t_n=2n-0.5$

A recursive rule for an AP is $t_1=a$, $t_n=t_{n-1}+d$ for $n\ge 2$:

• 2, 5, 8, 11, ...: $t_1=2$, $t_n=t_{n-1}+3$
• −5, −1, 3, 7, ...: $t_1=-5$, $t_n=t_{n-1}+4$
• $\frac{1}{2},\frac{5}{2},\frac{9}{2},\dots$: $t_1=\frac{1}{2}$, $t_n=t_{n-1}+2$
• 1.5, 3.5, 5.5, ...: $t_1=1.5$, $t_n=t_{n-1}+2$

Start with the first term and add the common difference each time, as listed above.

a = 3, d = 5.

$$t_{10}=3+9\times 5=48,t_{26}=3+25\times 5=128$$

t₁₀ = 48 and t₂₆ = 128

a = 21, d = −3, so $t_n=21-3(n-1)=24-3n$.

$$24-3n=-81\Rightarrow 3n=105\Rightarrow n=35$$

$$24-3n=0\Rightarrow n=8$$

n = 8 is a natural number, so 0 is a term: 21, 18, 15, 12, 9, 6, 3, 0.

−81 is the 35th term; yes, 0 is the 8th term.

a = 11, d = −3:

$$t_n=11-3(n-1)=14-3n$$

Recursive rule: $t_1=11$, $t_n=t_{n-1}-3$ for $n\ge 2$.

tₙ = 14 − 3n; t₁ = 11, tₙ = tₙ₋₁ − 3.

The last term is the 50th term. So

$$a+2d=12\text{and}a+49d=106$$

Subtracting: $47d=94$, so $d=2$ and $a=12-4=8$.

$$t_{29}=8+28\times 2=64$$

The 29th term is 64.

They form the AP 12, 15, 18, ..., 99 with d = 3.

$$99=12+(n-1)\times 3\Rightarrow n-1=29\Rightarrow n=30$$

Sum = number of terms × average of first and last term:

$$S=\frac{30}{2}(12+99)=15\times 111=1665$$

30 numbers; their sum is 1665.

The salaries form an AP with a = 500000, d = 20000:

$$500000+20000(n-1)=700000\Rightarrow n-1=10\Rightarrow n=11$$

His salary is ₹7,00,000 in the 11th year, i.e. after 10 increments.

After 10 years (his 11th year of work), his income reached ₹7,00,000.

$$1+2+\cdots +25=\frac{25\times 26}{2}=325$$

325 marbles.

Check the ratio of consecutive terms.

(i) $\frac{10}{2}=\frac{50}{10}=\frac{250}{50}=5$. A GP with a = 2, r = 5: $t_n=2\times 5^{n-1}$.

(ii) $\frac{8/3}{4}=\frac{16/9}{8/3}=\frac{32/27}{16/9}=\frac{2}{3}$. A GP with a = 4, $r=\frac{2}{3}$: $t_n=4{\left(\frac{2}{3}\right)}^{n-1}$.

(iii) $\frac{-3/2}{3}=\frac{3/4}{-3/2}=\frac{-3/8}{3/4}=-\frac{1}{2}$. A GP with a = 3, $r=-\frac{1}{2}$: $t_n=3{\left(-\frac{1}{2}\right)}^{n-1}$.

All three are GPs: (i) $2\times 5^{n-1}$ (ii) $4{\left(\frac{2}{3}\right)}^{n-1}$ (iii) $3{\left(-\frac{1}{2}\right)}^{n-1}$

Each term is 10 times the previous one: $t_1=3$, $t_n=10t_{n-1}$ for $n\ge 2$.

t₁ = 3, tₙ = 10 tₙ₋₁ (n ≥ 2)

Going from the 8th term to the 12th term multiplies by r four times:

$$t_{12}=t_8\times r^4=192\times 16=3072$$

(Also $a=\frac{192}{2^7}=1.5$, and $1.5\times 2^{11}=3072$ ✓.)

The 12th term is 3072.

a = 5, r = 5, so $t_n=5\times 5^{n-1}=5^n$ and $t_{10}=5^{10}=9765625$.

t₁₀ = 9765625; tₙ = 5ⁿ

Computing terms: 2, 4, 10, 28, 82, 244, 730. So 730 is the 7th term.

A neater way: subtract 1 from each term to get 1, 3, 9, 27, 81, 243, 729, which is a GP: $t_n-1=3^{n-1}$, so $t_n=3^{n-1}+1$. Then $3^{n-1}=729=3^6$ gives $n=7$.

730 is the 7th term.

a = 2, r = 3, so $t_n=2\times 3^{n-1}$.

$$2\times 3^{n-1}=4374\Rightarrow 3^{n-1}=2187=3^7\Rightarrow n=8$$

Recursive formula: $t_1=2$, $t_n=3t_{n-1}$ for $n\ge 2$.

4374 is the 8th term; tₙ = 2 × 3ⁿ⁻¹; t₁ = 2, tₙ = 3tₙ₋₁.

The bounce heights form a GP with first term $80\times 0.6=48$ and r = 0.6: 48, 28.8, 17.28, 10.368, 6.2208, ...

(i) After the 5th bounce: $80\times 0.6^5=6.2208$ m.

(ii) The ball falls 80 m and hits the ground (1st time). After each of the next 5 bounces it goes up and comes down the same height before hitting the ground again (2nd to 6th times).

$$\text{Distance}=80+2(48+28.8+17.28+10.368+6.2208)=80+2\times 110.6688=301.3376\text{m}$$

(i) 6.2208 m (ii) about 301.34 m.

It is a GP with a = 2 and $r=\sqrt{2}$. Write everything as powers of 2: $t_n=2\times (\sqrt{2}{)}^{n-1}=2^{1+\frac{n-1}{2}}$ and $128=2^7$.

$$1+\frac{n-1}{2}=7\Rightarrow n-1=12\Rightarrow n=13$$

128 is the 13th term.

Each red square is replaced by 8 smaller red squares, each with $\frac{1}{9}$ of its area.

(i) Stages 0 to 3: 1, 8, 64, 512.

(ii) Stage 4: $512\times 8=4096$; Stage 5: $4096\times 8=32768$.

(iii) Explicit: $t_n=8^n$. Recursive: $t_0=1$, $t_n=8t_{n-1}$ for $n\ge 1$.

(iv) At each stage the red area is multiplied by $\frac{8}{9}$:

Explicit: $s_n={\left(\frac{8}{9}\right)}^n$. Recursive: $s_0=1$, $s_n=\frac{8}{9}{s}_{n-1}$ for $n\ge 1$. As n increases, the number of squares grows very fast, but the red area keeps shrinking and gets closer and closer to 0.

Squares: 1, 8, 64, 512, then 4096, 32768; tₙ = 8ⁿ. Area: sₙ = (8/9)ⁿ, which tends to 0 as n increases.

$$a+10d=38,a+15d=73\Rightarrow 5d=35\Rightarrow d=7,a=38-70=-32$$

$$t_{31}=-32+30\times 7=178$$

The 31st term is 178.

$t_7-t_5=2d=12$, so d = 6. Then $a+2d=16$ gives a = 4.

The AP is 4, 10, 16, 22, 28, ...

The smallest is 105 (= 7 × 15) and the largest is 994 (= 7 × 142). They form an AP with d = 7:

$$994=105+(n-1)\times 7\Rightarrow n-1=127\Rightarrow n=128$$

128 three-digit numbers are divisible by 7.

The smallest multiple of 4 above 10 is 12 and the largest below 250 is 248:

$$248=12+(n-1)\times 4\Rightarrow n-1=59\Rightarrow n=60$$

60 multiples of 4.

$$a{r}^4=4a{r}^2\Rightarrow r^2=4\Rightarrow r=2\text{ or }r=-2$$

• r = 2: $a+2a=-4$, so $a=-\frac{4}{3}$: the GP $-\frac{4}{3},-\frac{8}{3},-\frac{16}{3},\dots$
• r = −2: $a-2a=-4$, so $a=4$: the GP 4, −8, 16, −32, ...

Either 4, −8, 16, −32, ... (r = −2) or −4/3, −8/3, −16/3, ... (r = 2).

Suppose k consecutive numbers starting from a add up to 100. Their sum is $\frac{k}{2}(2a+k-1)=100$, so

$$k(2a+k-1)=200$$

The two factors k and $2a+k-1$ differ by an odd number ($2a-1$), so one is odd and the other even, and k is the smaller. The factor pairs of 200 of this type are 1 × 200, 5 × 40 and 8 × 25.

• k = 1: just 100 itself.
• k = 5: $2a+4=40$, a = 18: $18+19+20+21+22=100$.
• k = 8: $2a+7=25$, a = 9: $9+10+11+12+13+14+15+16=100$.

Two ways (using two or more numbers): 18 + 19 + 20 + 21 + 22 and 9 + 10 + 11 + ... + 16.

After n hours: $30\times 2^n$.

$$\text{2nd hour: }30\times 4=120,\text{4th hour: }30\times 16=480$$

120, 480 and 30 × 2ⁿ.

$$(a+3d)+(a+7d)=24\Rightarrow 2a+10d=24$$

$$(a+5d)+(a+9d)=44\Rightarrow 2a+14d=44$$

Subtracting: $4d=20$, so d = 5 and $2a=24-50=-26$, a = −13.

The first three terms are −13, −8, −3.

We need $\frac{n(n+1)}{2}>1000$, i.e. $n(n+1)>2000$. Since $44\times 45=1980$ and $45\times 46=2070$:

$$S_{44}=990<1000,S_{45}=1035>1000$$

n = 45

a = 2, r = 4:

$$2\times 4^{n-1}=131072\Rightarrow 4^{n-1}=65536=4^8\Rightarrow n=9$$

Explicit: $t_n=2\times 4^{n-1}$. Recursive: $t_1=2$, $t_n=4t_{n-1}$ for $n\ge 2$.

131072 is the 9th term; tₙ = 2 × 4ⁿ⁻¹; t₁ = 2, tₙ = 4tₙ₋₁.

Take the three terms as $\frac{a}{r},a,ar$. Their product is $a^3=-1$, so $a=-1$. Then

$$-\frac{1}{r}-1-r=\frac{13}{12}\Rightarrow r+\frac{1}{r}=-\frac{25}{12}\Rightarrow 12r^2+25r+12=0$$

$$(4r+3)(3r+4)=0\Rightarrow r=-\frac{3}{4}\text{ or }r=-\frac{4}{3}$$

With $r=-\frac{3}{4}$ the terms are $\frac{4}{3},-1,\frac{3}{4}$; with $r=-\frac{4}{3}$ they are the same numbers in reverse order. Check: $\frac{4}{3}-1+\frac{3}{4}=\frac{16-12+9}{12}=\frac{13}{12}$ ✓ and $\frac{4}{3}\times (-1)\times \frac{3}{4}=-1$ ✓

r = −3/4 with terms 4/3, −1, 3/4 (or r = −4/3 with terms 3/4, −1, 4/3).

$$x=a{r}^3,y=a{r}^9,z=a{r}^{15}$$

$$\frac{y}{x}=r^6\text{and}\frac{z}{y}=r^6$$

The ratios are equal (equivalently, $y^2=a^2{r}^{18}=xz$), so x, y, z are in GP with common ratio $r^6$.

y/x = z/y = r⁶, so x, y, z form a GP.

Let the terms be $a,ar,a{r}^2$:

$$a(1+r+r^2)=26,a^2(1+r^2+r^4)=364$$

Use $1+r^2+r^4=(1+r+r^2)(1-r+r^2)$. Dividing the second equation by the first:

$$a(1-r+r^2)=\frac{364}{26}=14$$

Subtracting this from $a(1+r+r^2)=26$: $2ar=12$, so $ar=6$. Then $a+a{r}^2=26-6=20$, i.e. $a+\frac{36}{a}=20$:

$$a^2-20a+36=0\Rightarrow (a-2)(a-18)=0$$

So a = 2 (r = 3) or a = 18 (r = 1/3). Check: $2+6+18=26$ and $4+36+324=364$ ✓

The terms are 2, 6, 18 (or 18, 6, 2).

$$P_3=1+2+1=4,P_4=1+2+4+1=8,P_5=16,P_6=32,P_7=64,P_8=128$$

Simpler rule: $P_n=(P_1+\cdots +P_{n-2}+1)+P_{n-1}=P_{n-1}+P_{n-1}=2P_{n-1}$ (for n > 3, and it also holds for n = 2, 3). Explicit formula: $P_n=2^{n-1}$.

1, 2, 4, 8, 16, 32, 64, 128; Pₙ = 2Pₙ₋₁; Pₙ = 2ⁿ⁻¹.

$$W_3=1+2=3,W_4=1+2+2=5,W_5=1+2+3+2=8,W_6=13,W_7=21,W_8=34$$

The terms 1, 2, 3, 5, 8, 13, 21, 34 are the Virahānka–Fibonacci numbers. Indeed, $W_n-W_{n-1}=W_{n-2}$, so $W_n=W_{n-1}+W_{n-2}$.

1, 2, 3, 5, 8, 13, 21, 34: the Virahānka–Fibonacci sequence.