What is the formula for the nth term of an AP?

The nth term equals the first term a plus (n minus 1) times the common difference d, where d is the constant amount added to get from one term to the next.

What is the formula for the nth term of a GP?

The nth term equals the first term a times r to the power (n minus 1), where r is the common ratio, the constant factor multiplied to get from one term to the next.

What is the Virahanka-Fibonacci sequence?

It is the sequence 1, 2, 3, 5, 8, 13, 21, 34 and so on, where each term is the sum of the previous two. It was first written down by the Indian scholar Virahanka in the 7th century CE while studying Prakrit poetic meter, centuries before it was studied in Europe by Fibonacci.

Class 9 Maths Ganita Manjari NCERT Solutions Chapter 8 - Predicting What Comes Next: Sequences and Progressions

Key Concepts & Formulae at a Glance

A sequence is an ordered list of numbers; each number is called a term. A sequence may be finite or infinite.
An explicit formula gives the value of a term directly from its position number n.
A recursive formula gives a term's value using the value(s) of previous term(s).
An arithmetic progression (AP) has a constant common difference d between consecutive terms.
A geometric progression (GP) has a constant common ratio r between consecutive terms.
The Virahānka–Fibonacci sequence (1, 1, 2, 3, 5, 8, 13, ...) is generated by adding the previous two terms — first studied by Virahānka in the 7th century CE.

\[ \text{Triangular numbers: } t_n = \frac{n(n+1)}{2} \quad\text{(also the sum of the first } n \text{ natural numbers, } S_n\text{)} \] \[ \text{AP: } t_n = a + (n-1)d, \quad \text{general form } a,\ a+d,\ a+2d,\ \ldots \] \[ \text{GP: } t_n = ar^{n-1}, \quad \text{general form } a,\ ar,\ ar^2,\ \ldots \]

TRCan you describe the pattern in each of the sequences 1,2,3,4,5,6,…; 1,3,5,7,9,11,…; 1,3,6,10,15,21,…; 1,4,9,16,25,36,…? Can you predict the next few numbers in these sequences?

Natural numbers 1,2,3,4,5,6,…: each term is 1 more than the previous term. Next terms: 7, 8, 9.

Odd numbers 1,3,5,7,9,11,…: each term is 2 more than the previous term. Next terms: 13, 15, 17.

Triangular numbers 1,3,6,10,15,21,…: the difference between consecutive terms increases by 1 each time (2,3,4,5,6,...); each term is the sum of natural numbers up to that position. Next terms: \(21+7=28\), \(28+8=36\), \(36+9=45\).

Square numbers 1,4,9,16,25,36,…: the nth term is n². Next terms: \(7^2=49\), \(8^2=64\), \(9^2=81\).

Natural: +7,8,9. Odd: +13,15,17. Triangular: +28,36,45. Square: +49,64,81.

ExExercise: Consider the sequence 1, 4, 7, 10, 13, … Can you predict the next four terms? Can you derive the first 10 terms of the sequence obtained by adding all the terms up to a given term of this sequence?

The sequence increases by 3 each time (common difference 3), so the next four terms are: \(13+3=16\), \(19\), \(22\), \(25\).

First 10 terms of 1,4,7,10,13,…: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28

Partial sums (running total up to each term):

\(1;\)
\(1+4=5;\)
\(5+7=12;\)
\(12+10=22;\)
\(22+13=35;\)
\(35+16=51;\)
\(51+19=70;\)
\(70+22=92;\)
\(92+25=117;\)
\(117+28=145\)

Next four terms: 16, 19, 22, 25. Partial-sum sequence (first 10 terms): 1, 5, 12, 22, 35, 51, 70, 92, 117, 145.

ExExercise: Can you write t₅, t₆, t₇ and t₈ for the sequence of triangular numbers?

Triangular numbers: 1, 3, 6, 10, 15, 21, 28, 36, ...

t₅ = 15, t₆ = 21, t₇ = 28, t₈ = 36.

TRCan you think of any other kinds of sequences? List out five different types of sequences and discuss their properties with your friends.

Some other kinds of sequences, beyond the ones already discussed:

1. Prime numbers: 2, 3, 5, 7, 11, 13, … — no simple explicit formula is known; the gaps between terms are irregular.

2. Cube numbers: 1, 8, 27, 64, 125, … — explicit formula \(t_n=n^3\).

3. Geometric sequences: e.g. 1, 2, 4, 8, 16, … — each term is a constant multiple of the previous one.

4. Alternating sequences: e.g. 1, −1, 1, −1, … — terms alternate in sign, never settling into pure increase or decrease.

5. Decreasing fraction sequences: e.g. 1, 1/2, 1/3, 1/4, … — terms get smaller and approach 0 but never reach it.

Examples: primes (no simple formula), cubes (tₙ = n³), geometric sequences (constant ratio), alternating sequences (changing sign), and decreasing unit fractions (approaching 0).

TRWhy is it useful to have an explicit formula for the nth term of a sequence?

With an explicit formula, we can find any term directly — the 20th, the 53rd, the 300th, or any other term — just by substituting the appropriate value of n.

We don't need to know the value of any earlier terms first, unlike with a recursive rule, where each term depends on knowing the one(s) before it.

An explicit formula lets us compute any term directly from its position n, without first calculating all the earlier terms.

ExExercise: Using the explicit rule uₙ = 2n − 1, find the 53rd term, the 108th term, and the 1170th term of the odd number sequence.

\(u_{53}=2(53)-1=105\)

\(u_{108}=2(108)-1=215\)

\(u_{1170}=2(1170)-1=2339\)

53rd term = 105, 108th term = 215, 1170th term = 2339.

TRCan you find the rule describing the nth term of the sequence of square numbers?

The square number sequence is 1, 4, 9, 16, 25, …, where each term is the square of its position number.

tₙ = n².

ExExercise: Consider the expression tₙ = 3n − 7.
(i) Find its first, second, third, 12th, 18th and 50th terms.
(ii) Which term of the sequence is 332?
(iii) Is 557 a term of this sequence? Why or why not?

(i) \(t_1=3(1)-7=-4\);
\(t_2=3(2)-7=-1\);
\(t_3=3(3)-7=2\);
\(t_{12}=3(12)-7=29\);
\(t_{18}=3(18)-7=47\);
\(t_{50}=3(50)-7=143\)

(ii) \(3n-7=332 \Rightarrow 3n=339 \Rightarrow n=113\). So 332 is the 113th term.

(iii) \(3n-7=557 \Rightarrow 3n=564 \Rightarrow n=188\). Since 188 is a natural number, yes, 557 is a term of this sequence (the 188th term).

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

ExThe Virahānka–Fibonacci sequence: V₁ = 1, V₂ = 2, Vₙ = Vₙ₋₁ + Vₙ₋₂ for n ≥ 3, giving 1, 2, 3, 5, 8, 13, 21, 34, … Can you write the next two terms of this sequence?

Each term is the sum of the previous two: \(34+21=55\), then \(55+34=89\).

Next two terms: 55, 89.

Exercise Set 8.1

1Find the first five terms of the sequence in which the nth term is given by
(i) tₙ = 3n − 4,
(ii) tₙ = 2 − 5n, and
(iii) tₙ = n² − 2n + 3 for n ≥ 1.

(i) \(t_1=-1,\ t_2=2,\ t_3=5,\ t_4=8,\ t_5=11\)

(ii) \(t_1=-3,\ t_2=-8,\ t_3=-13,\ t_4=-18,\ t_5=-23\)

(iii) \(t_1=1-2+3=2\);
\(t_2=4-4+3=3\);
\(t_3=9-6+3=6\);
\(t_4=16-8+3=11\);
\(t_5=25-10+3=18\)

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

2Find the 10th and 15th terms of the sequence tₙ = 5n − 3 for n ≥ 1.

\(t_{10}=5(10)-3=47\)

\(t_{15}=5(15)-3=72\)

t₁₀ = 47, t₁₅ = 72.

3Determine whether 97 and 172 are terms of the sequence tₙ = 5n − 3 for n ≥ 1.

\(5n-3=97 \Rightarrow 5n=100 \Rightarrow n=20\) — a natural number, so 97 is a term (the 20th term).

\(5n-3=172 \Rightarrow 5n=175 \Rightarrow n=35\) — a natural number, so 172 is a term (the 35th term).

Both 97 (20th term) and 172 (35th term) are terms of the sequence.

4Which term of the sequence tₙ = 5n − 3 for n ≥ 1 is 607?

\(5n-3=607 \Rightarrow 5n=610 \Rightarrow n=122\)

607 is the 122nd term.

5A sequence is given by the recursive rule t₁ = −5, tₙ₊₁ = tₙ + 3 for n ≥ 1. Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?

\(t_1=-5,\ t_2=-5+3=-2,\ t_3=-2+3=1,\ t_4=1+3=4,\ t_5=4+3=7\)

This is an AP with \(a=-5, d=3\), so \(t_n=-5+(n-1)(3)=3n-8\).

Set \(3n-8=52 \Rightarrow 3n=60 \Rightarrow n=20\) — a natural number.

First five terms: −5, −2, 1, 4, 7. Yes, 52 is the 20th term.

6Let T₁ = 1, T₂ = 2, T₃ = 4, and Tₙ = Tₙ₋₁ + Tₙ₋₂ + Tₙ₋₃ for n ≥ 4. Find T₄, T₅, T₆, T₇, and T₈.

\(T_4=T_3+T_2+T_1=4+2+1=7\)

\(T_5=T_4+T_3+T_2=7+4+2=13\)

\(T_6=T_5+T_4+T_3=13+7+4=24\)

\(T_7=T_6+T_5+T_4=24+13+7=44\)

\(T_8=T_7+T_6+T_5=44+24+13=81\)

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

TRCan you predict the number of squares in Stages 5 and 6 of the growing pattern 1, 5, 9, 13, …? In Stages 10, 11 and 12? In Stage 20? At any stage?

4 squares are added at each stage — 1, 5, 9, 13, ... (constant difference 4, an AP)

This sequence is an AP with \(a=1, d=4\), so \(t_n=1+(n-1)(4)=4n-3\).

Stage 5: \(4(5)-3=17\); Stage 6: \(4(6)-3=21\)

Stage 10: \(4(10)-3=37\);
Stage 11: \(4(11)-3=41\);
Stage 12: \(4(12)-3=45\)

Stage 20: \(4(20)-3=77\)

Stage 5 = 17, Stage 6 = 21, Stage 10 = 37, Stage 11 = 41, Stage 12 = 45, Stage 20 = 77. At any stage n: 4n − 3.

TRConsider all the sequences discussed so far in this chapter. Which ones are arithmetic progressions and which ones are not? Can you justify your claim?

Are APs (constant difference between consecutive terms):

• Natural numbers 1,2,3,… (d = 1) • Odd numbers 1,3,5,… (d = 2) • 1,4,7,10,13,… (d = 3) • −7,−3,1,5,9,… (d = 4) • Growing-square-pattern 1,5,9,13,… (d = 4)

Are NOT APs (difference between consecutive terms is not constant):

• Triangular numbers 1,3,6,10,15,… (differences 2,3,4,5,… — increasing, not constant) • Square numbers 1,4,9,16,… (differences 3,5,7,9,… — increasing, not constant) • Prime numbers 2,3,5,7,11,… (differences 1,2,2,4,… — irregular) • Unit fractions 1,1/2,1/3,1/4,… (differences shrink, not constant)

APs: natural numbers, odd numbers, 1,4,7,10,13,..., −7,−3,1,5,9,..., and the growing-square pattern. Not APs: triangular numbers, square numbers, primes, and unit fractions — each has a non-constant difference between consecutive terms.

ExExercise: Verify that the following sequences are arithmetic progressions and write their nth terms. What do you observe when you plot the ordered pairs emerging from them?
(i) 2, 5, 8, 11, …
(ii) −5, −1, 3, 7, …

(i) Differences: \(5-2=3,\ 8-5=3,\ 11-8=3\) — constant, so it's an AP with \(a=2,d=3\). \(t_n=2+(n-1)(3)=3n-1\)

(ii) Differences: \(-1-(-5)=4,\ 3-(-1)=4,\ 7-3=4\) — constant, so it's an AP with \(a=-5,d=4\). \(t_n=-5+(n-1)(4)=4n-9\)

Since \(t_n\) is a linear expression in n in both cases, the plotted points \((n,t_n)\) all lie on a straight line.

Both sequences plotted as (n, tₙ) — each set of points lies on a straight line

(i) AP, tₙ = 3n − 1 (ii) AP, tₙ = 4n − 9. In both cases, the plotted (n, tₙ) points lie on a straight line, since tₙ is linear in n.

ExExercise: Using the formula tₙ = a + (n − 1) × d, find the nth term of the following arithmetic progressions.
(i) 1/2, 5/2, 9/2, 13/2, …
(ii) 1.5, 3.5, 5.5, 7.5, …

(i) \(a=\dfrac{1}{2}, d=\dfrac{5}{2}-\dfrac{1}{2}=2\). \(t_n=\dfrac{1}{2}+(n-1)(2)=2n-\dfrac{3}{2}\)

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

(i) tₙ = \(2n-\dfrac{3}{2}\) (ii) tₙ = 2n − 0.5.

ExExercise: Find recursive rules for the APs in the previous exercises.

2, 5, 8, 11, …: \(t_1=2,\ t_n=t_{n-1}+3\) for \(n\ge2\)

−5, −1, 3, 7, …: \(t_1=-5,\ t_n=t_{n-1}+4\) for \(n\ge2\)

1/2, 5/2, 9/2, 13/2, …: \(t_1=\dfrac12,\ t_n=t_{n-1}+2\) for \(n\ge2\)

1.5, 3.5, 5.5, 7.5, …: \(t_1=1.5,\ t_n=t_{n-1}+2\) for \(n\ge2\)

Each recursive rule is t₁ = (first term), tₙ = tₙ₋₁ + d for n ≥ 2, using the appropriate first term and common difference d for each sequence.

TRCan the same approach (writing the sum forwards and backwards, then adding) be used to find the sum of 1 + 2 + 3 + ⋯ + 100?

Yes. Let \(S=1+2+\cdots+100\). Writing it backwards: \(S=100+99+\cdots+1\).

Adding both: \(2S=101+101+\cdots+101\) (101 added 100 times) \(=101\times100=10100\)

\(S=\dfrac{10100}{2}=5050\)

Yes — 1 + 2 + ⋯ + 100 = 5050, using the same forward/backward pairing method.

TRCan you use the formula Sₙ = n(n+1)/2 to find S₂₀, S₅₀ or S₁₀₀₀?

\(S_{20}=\dfrac{20\times21}{2}=210\)

\(S_{50}=\dfrac{50\times51}{2}=1275\)

\(S_{1000}=\dfrac{1000\times1001}{2}=500500\)

S₂₀ = 210, S₅₀ = 1275, S₁₀₀₀ = 500500.

TRThe nth term of the triangular number sequence is tₙ = n(n+1)/2. Can you use this to find the 10th, 17th and 80th triangular numbers?

\(t_{10}=\dfrac{10\times11}{2}=55\)

\(t_{17}=\dfrac{17\times18}{2}=153\)

\(t_{80}=\dfrac{80\times81}{2}=3240\)

10th triangular number = 55, 17th = 153, 80th = 3240.

Exercise Set 8.2

1Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ….

\(a=3, d=5\)

\(t_{10}=3+(10-1)(5)=3+45=48\)

\(t_{26}=3+(26-1)(5)=3+125=128\)

t₁₀ = 48, t₂₆ = 128.

2Which term of the AP: 21, 18, 15, … is −81? Also, is 0 a term of this AP? Give reasons for your answer.

\(a=21, d=-3\). \(t_n=21+(n-1)(-3)=24-3n\)

\(24-3n=-81 \Rightarrow -3n=-105 \Rightarrow n=35\)

Is 0 a term? \(24-3n=0 \Rightarrow n=8\) — a natural number, so yes, 0 is the 8th term.

−81 is the 35th term. Yes, 0 is also a term — the 8th term.

3Find the nth term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.

\(a=11, d=-3\). \(t_n=11+(n-1)(-3)=14-3n\)

Recursive rule: \(t_1=11,\ t_n=t_{n-1}-3\) for \(n\ge2\)

tₙ = 14 − 3n. Recursive: t₁ = 11, tₙ = tₙ₋₁ − 3 for n ≥ 2.

4An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.

\(a+2d=12\) …(1)

\(a+49d=106\) …(2)

(2) − (1): \(47d=94 \Rightarrow d=2\)

\(a=12-2(2)=8\)

\(t_{29}=a+28d=8+28(2)=8+56=64\)

29th term = 64.

5How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?

Smallest 2-digit multiple of 3: 12. Largest: 99. This is an AP: 12, 15, 18, …, 99, with \(a=12, d=3\).

\(12+(n-1)(3)=99 \Rightarrow (n-1)(3)=87 \Rightarrow n=30\)

Sum \(=\dfrac{n}{2}(a+l)=\dfrac{30}{2}(12+99)=15\times111=1665\)

There are 30 such numbers; their sum is 1665.

6Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000?

Salary forms an AP with \(a=5{,}00{,}000\) (starting salary) and \(d=20{,}000\) (yearly increment).

We need the number of increments k such that \(5{,}00{,}000+k(20{,}000)=7{,}00{,}000\)

\(k(20{,}000)=2{,}00{,}000 \Rightarrow k=10\)

After 10 years (10 increments), Harish's income reached ₹7,00,000.

7A child arranges marbles in rows so that the first row has 1 marble, the second has 2 marbles, the third has 3, and so on up to 25 rows. How many marbles does the child use in all?

Total \(=1+2+3+\cdots+25 = S_{25}=\dfrac{25\times26}{2}=325\)

The child uses 325 marbles in all.

TRCan you predict the number of squares in Stages 5 and 6 of the GP pattern 3, 6, 12, 24, …? In Stages 10, 11 and 12? In Stage 20? How is this different from the growing pattern in Fig. 8.3 (the AP)?

Illustrative: the GP pattern doubles in count at every stage (3, 6, 12, 24, ...)

This is a GP with \(a=3, r=2\), so \(t_n=3\times2^{n-1}\).

Stage 5: \(3\times2^4=48\); Stage 6: \(3\times2^5=96\)

Stage 10: \(3\times2^9=1536\);
Stage 11: \(3\times2^{10}=3072\);
Stage 12: \(3\times2^{11}=6144\)

Stage 20: \(3\times2^{19}=1{,}572{,}864\)

This differs from the AP pattern because the GP grows multiplicatively (doubling each stage), so its values increase far more rapidly than the AP, which grows by a fixed additive amount (4) at each stage.

Stage 5 = 48, Stage 6 = 96, Stage 10 = 1536, Stage 11 = 3072, Stage 12 = 6144, Stage 20 = 1,572,864. Unlike the AP (constant additive growth), the GP grows multiplicatively — far faster.

ExExamples: Is 1, 2, 4, 8, 16, … a geometric progression? Is 1, 3, 9, 27, 81, … a geometric progression? Is 1, −1, 1, −1, 1, … a geometric progression? In each case, if so, what is the common ratio?

1, 2, 4, 8, 16, …: ratios \(\dfrac{2}{1}=2,\ \dfrac{4}{2}=2,\ \dfrac{8}{4}=2,\ \dfrac{16}{8}=2\) — constant. Yes, GP with r = 2.

1, 3, 9, 27, 81, …: ratios are all 3. Yes, GP with r = 3.

1, −1, 1, −1, 1, …: ratios are all −1. Yes, GP with r = −1.

All three are GPs, with common ratios 2, 3, and −1 respectively.

ExExercise: Check whether the following sequences are geometric progressions and find their nth terms.
(i) 2, 10, 50, 250, …
(ii) 4, 8/3, 16/9, 32/27, …
(iii) 3, −3/2, 3/4, −3/8, …

(i) Ratios: \(\dfrac{10}{2}=5,\ \dfrac{50}{10}=5,\ \dfrac{250}{50}=5\) — constant. GP with \(a=2, r=5\). \(t_n=2\times5^{n-1}\)

(ii) Ratios: \(\dfrac{8/3}{4}=\dfrac23,\ \dfrac{16/9}{8/3}=\dfrac23,\ \dfrac{32/27}{16/9}=\dfrac23\) — constant. GP with \(a=4, r=\dfrac23\). \(t_n=4\times\left(\dfrac23\right)^{n-1}\)

(iii) Ratios: \(\dfrac{-3/2}{3}=-\dfrac12,\ \dfrac{3/4}{-3/2}=-\dfrac12,\ \dfrac{-3/8}{3/4}=-\dfrac12\) — constant. GP with \(a=3, r=-\dfrac12\). \(t_n=3\times\left(-\dfrac12\right)^{n-1}\)

All three are GPs: (i) tₙ = 2×5ⁿ⁻¹ (ii) tₙ = 4×\(\left(\dfrac{2}{3}\right)^{n-1}\) (iii) tₙ = 3×\(\left(-\dfrac{1}{2}\right)^{n-1}\).

ExExercise: Can you find a recursive rule for the formula tₙ = 3 × 10ⁿ⁻¹ that generates the geometric progression 3, 30, 300, 3000, … ?

Since each term is 10 times the previous term:

t₁ = 3, tₙ = 10 × tₙ₋₁ for n ≥ 2.

TRObserve the Sierpiński triangle:
(a) How many black triangles are there in Stages 0 to 3?
(b) Predict the number of black triangles at Stages 4 and 5.
(c) Find a rule for the number of black triangles at the nth stage.
(d) Suppose the area of the black region in Stage 0 is 1 square unit. Find the area of the black region in Stages 1, 2 and 3, and predict Stages 4 and 5. Find a rule for the area at the nth stage. What happens to this area as n increases?

Sierpiński triangle, Stages 0-3 — 1, 3, 9, 27 black triangles (drawn to exact recursive scale)

(a) Stages 0–3: 1, 3, 9, 27 black triangles.

(b) Each stage has 3 times as many black triangles as the previous stage (every black triangle splits into 3 smaller ones). Stage 4: \(27\times3=81\); Stage 5: \(81\times3=243\)

(c) The sequence 1, 3, 9, 27, 81, 243, … is a GP with \(a=1, r=3\), so the number of black triangles at stage n is \(3^n\).

(d) At each stage, the black region is multiplied by \(\dfrac34\) (since the central quarter is removed each time).
Stage 1: \(\dfrac34\);
Stage 2: \(\left(\dfrac34\right)^2=\dfrac{9}{16}\);
Stage 3: \(\left(\dfrac34\right)^3=\dfrac{27}{64}\)

Predicted: Stage 4: \(\left(\dfrac34\right)^4=\dfrac{81}{256}\); Stage 5: \(\left(\dfrac34\right)^5=\dfrac{243}{1024}\)

Rule for stage n: area \(=\left(\dfrac34\right)^n\). Since \(\dfrac34<1\), this area gets smaller and smaller, approaching 0 as n increases (though it never actually reaches 0).

(a) 1, 3, 9, 27 (b) 81, 243 (c) 3ⁿ black triangles at stage n (d) Areas: \(\dfrac{3}{4}, \dfrac{9}{16}, \dfrac{27}{64}\) (Stages 1–3), \(\dfrac{81}{256}, \dfrac{243}{1024}\) (Stages 4–5); rule \(\left(\dfrac{3}{4}\right)^n\); the area approaches 0 as n increases.

Exercise Set 8.3

1Find the 12th term of a GP with common ratio 2, whose 8th term is 192.

\(t_{12}=t_8\times r^{(12-8)}=192\times2^4=192\times16=3072\)

12th term = 3072.

2Find the 10th and nth terms of the GP: 5, 25, 125, … .

\(a=5, r=5\). \(t_n=5\times5^{n-1}=5^n\)

\(t_{10}=5^{10}=9{,}765{,}625\)

10th term = 9,765,625. nth term: tₙ = 5ⁿ.

3*A sequence is given by the recursive rule t₁ = 2, tₙ₊₁ = 3tₙ − 2 for n ≥ 1. Which term of the sequence is 730?

\(t_1=2\)

\(t_2=3(2)-2=4\)

\(t_3=3(4)-2=10\)

\(t_4=3(10)-2=28\)

\(t_5=3(28)-2=82\)

\(t_6=3(82)-2=244\)

\(t_7=3(244)-2=730\)

730 is the 7th term.

4Which term of the GP: 2, 6, 18, … is 4374? Write the explicit formula as well as the recursive formula for the nth term.

\(a=2, r=3\). \(t_n=2\times3^{n-1}=4374 \Rightarrow 3^{n-1}=2187=3^7 \Rightarrow n-1=7 \Rightarrow n=8\)

Explicit formula: \(t_n=2\times3^{n-1}\)

Recursive formula: \(t_1=2,\ t_n=3\times t_{n-1}\) for \(n\ge2\)

4374 is the 8th term. Explicit: tₙ = 2×3ⁿ⁻¹. Recursive: t₁ = 2, tₙ = 3tₙ₋₁ for n ≥ 2.

5A ball is dropped from a height of 80 metres. After hitting the ground, it bounces back to 60% of the height from which it fell, and continues bouncing this way.
(i) What height does the ball reach after the 5th bounce?
(ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?

(i) Bounce heights form a GP with \(a=80\times0.6=48\) (height after 1st bounce) and \(r=0.6\).

\(h_5=80\times(0.6)^5=80\times0.07776=6.2208\text{ m}\)

(ii) By the 6th ground-hit, the ball has: fallen the initial 80 m, then bounced up-and-down 5 times (for hits 2 through 6), using heights \(h_1,\ldots,h_5=48,\ 28.8,\ 17.28,\ 10.368,\ 6.2208\)

Sum of these 5 bounce heights: \(48+28.8+17.28+10.368+6.2208=110.6688\)

Total distance \(=80+2\times110.6688=80+221.3376=301.3376\text{ m}\)

(i) ≈ 6.22 m after the 5th bounce. (ii) Total distance ≈ 301.34 m by the 6th ground-hit.

6Which term of the sequence 2, 2√2, 4, … is 128?

Ratio: \(\dfrac{2\sqrt2}{2}=\sqrt2\); \(\dfrac{4}{2\sqrt2}=\sqrt2\) — constant. This is a GP with \(a=2, r=\sqrt2\).

\(t_n=2\times(\sqrt2)^{n-1}=2^{1+\frac{n-1}{2}}=2^{\frac{n+1}{2}}\)

Set \(2^{\frac{n+1}{2}}=128=2^7\): \(\dfrac{n+1}{2}=7 \Rightarrow n=13\)

128 is the 13th term.

7Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet.
(i) How many red squares are there in Stages 0 to 3?
(ii) Predict the number of red squares in Stages 4 and 5.
(iii) Find a rule (explicit and recursive) for the number of red squares at the nth stage.
(iv) Suppose the area of the square in Stage 0 is 1 square unit. Find the area of the red region in Stages 1, 2 and 3, predict Stages 4 and 5, and find explicit and recursive formulas for the area at the nth stage. What happens to this area as n increases?

Fig. 8.12: Stages 0-3 of the Sierpiński square carpet (1, 8, 64, 512 red squares — drawn to exact scale)

(i) Stage 0: 1 (whole square). Each stage keeps 8 of the 9 smaller squares from the previous stage.
Stage 1: 8;
Stage 2: \(8\times8=64\);
Stage 3: \(64\times8=512\)

(ii) Stage 4: \(512\times8=4096\); Stage 5: \(4096\times8=32{,}768\)

(iii) This is a GP with \(a=1, r=8\) (indexing stages from 0). Explicit: \(t_n=8^n\). Recursive: \(t_0=1,\ t_n=8\times t_{n-1}\) for \(n\ge1\)

(iv) Each stage retains \(\dfrac89\) of the previous stage's area.
Stage 1: \(\dfrac89\);
Stage 2: \(\left(\dfrac89\right)^2=\dfrac{64}{81}\);
Stage 3: \(\left(\dfrac89\right)^3=\dfrac{512}{729}\)

Predicted: Stage 4: \(\left(\dfrac89\right)^4=\dfrac{4096}{6561}\); Stage 5: \(\left(\dfrac89\right)^5=\dfrac{32768}{59049}\)

Explicit formula: area \(=\left(\dfrac89\right)^n\). Recursive: \(s_0=1,\ s_n=\dfrac89\times s_{n-1}\) for \(n\ge1\). Since \(\dfrac89<1\), the area approaches 0 as n increases.

(i) 1, 8, 64, 512 (ii) 4096, 32768 (iii) tₙ = 8ⁿ (iv) Areas \(\dfrac{8}{9}, \dfrac{64}{81}, \dfrac{512}{729}\) (Stages 1–3); \(\dfrac{4096}{6561}, \dfrac{32768}{59049}\) (Stages 4–5); rule \(\left(\dfrac{8}{9}\right)^n\); area → 0 as n increases.

End-of-Chapter Exercises

1Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

\(a+10d=38\) …(1)

\(a+15d=73\) …(2)

(2) − (1): \(5d=35 \Rightarrow d=7\)

\(a=38-10(7)=-32\)

\(t_{31}=a+30d=-32+30(7)=-32+210=178\)

31st term = 178.

2Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.

\(t_3=a+2d=16\)

\(t_7-t_5=12 \Rightarrow (a+6d)-(a+4d)=12 \Rightarrow 2d=12 \Rightarrow d=6\)

\(a+2(6)=16 \Rightarrow a=4\)

The AP is 4, 10, 16, 22, 28, … (a = 4, d = 6).

3*How many three-digit numbers are divisible by 7?

Smallest 3-digit multiple of 7: \(100\div7\approx14.3\), so \(15\times7=105\). Largest: \(999\div7\approx142.7\), so \(142\times7=994\).

AP: 105, 112, …, 994, with \(a=105, d=7, l=994\).

\(105+(n-1)(7)=994 \Rightarrow (n-1)(7)=889 \Rightarrow n=128\)

There are 128 three-digit numbers divisible by 7.

4*How many multiples of 4 lie between 10 and 250?

Smallest multiple of 4 greater than 10: 12. Largest multiple of 4 less than 250: 248.

AP: 12, 16, …, 248, with \(a=12, d=4, l=248\).

\(12+(n-1)(4)=248 \Rightarrow (n-1)(4)=236 \Rightarrow n=60\)

There are 60 multiples of 4 between 10 and 250.

5*Find a GP for which the sum of the first two terms is −4 and the fifth term is 4 times the third term.

\(a+ar=-4 \Rightarrow a(1+r)=-4\) …(1)

\(ar^4=4(ar^2) \Rightarrow r^2=4 \Rightarrow r=\pm2\)

Case r = 2: \(a(3)=-4 \Rightarrow a=-\dfrac43\). GP: \(-\dfrac43,-\dfrac83,-\dfrac{16}{3},\ldots\)

Case r = −2: \(a(-1)=-4 \Rightarrow a=4\). GP: \(4,-8,16,-32,\ldots\)

Two GPs satisfy the conditions: 4, −8, 16, −32, ... (r = −2), or \(-\dfrac{4}{3}, -\dfrac{8}{3}, -\dfrac{16}{3}\), ... (r = 2).

6*Find all possible ways of expressing 100 as the sum of consecutive natural numbers.

Sum of k consecutive naturals starting at a: \(ka+\dfrac{k(k-1)}{2}=100 \Rightarrow k(2a+k-1)=200\)

Testing divisors k of 200 (with \(k\ge2\)) for a positive integer value of a:

\(k=5\): \(2a+4=40 \Rightarrow a=18\) — valid. Sequence: 18, 19, 20, 21, 22.

\(k=8\): \(2a+7=25 \Rightarrow a=9\) — valid. Sequence: 9, 10, …, 16.

All other divisors of 200 give a non-integer or non-positive value of a.

100 = 18+19+20+21+22 (5 terms) = 9+10+11+12+13+14+15+16 (8 terms).

7*The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the 2nd hour, 4th hour and nth hour?

This is a GP with \(a=30\) (initial count) and \(r=2\). Number at end of hour n \(=30\times2^n\).

End of 2nd hour: \(30\times2^2=120\)

End of 4th hour: \(30\times2^4=480\)

End of 2nd hour: 120. End of 4th hour: 480. End of nth hour: 30 × 2ⁿ.

8*The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

\(t_4+t_8=24 \Rightarrow (a+3d)+(a+7d)=24 \Rightarrow a+5d=12\) …(1)

\(t_6+t_{10}=44 \Rightarrow (a+5d)+(a+9d)=44 \Rightarrow a+7d=22\) …(2)

(2) − (1): \(2d=10 \Rightarrow d=5\)

\(a=12-5(5)=-13\)

First three terms: −13, −8, −3.

9*Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.

Need \(\dfrac{n(n+1)}{2}>1000 \Rightarrow n(n+1)>2000\)

\(n=44\): \(44\times45=1980\) — not enough.

\(n=45\): \(45\times46=2070>2000\) — works. (Check: \(S_{45}=1035>1000\), \(S_{44}=990\), not enough.)

The smallest such n is 45.

10*Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula for the nth term.

\(a=2, r=4\). \(t_n=2\times4^{n-1}=131072 \Rightarrow 4^{n-1}=65536=4^8 \Rightarrow n-1=8 \Rightarrow n=9\)

Explicit formula: \(t_n=2\times4^{n-1}\)

Recursive formula: \(t_1=2,\ t_n=4\times t_{n-1}\) for \(n\ge2\)

131072 is the 9th term. Explicit: tₙ = 2×4ⁿ⁻¹. Recursive: t₁ = 2, tₙ = 4tₙ₋₁ for n ≥ 2.

11*The sum of the first three terms of a GP is 13/12 and their product is −1. Find the common ratio and the terms.

Let the three terms be \(\dfrac{a}{r}, a, ar\).

Product: \(\dfrac{a}{r}\times a\times ar=a^3=-1 \Rightarrow a=-1\)

Sum: \(\dfrac{a}{r}+a+ar=\dfrac{13}{12}\). Substituting \(a=-1\): \(-\dfrac1r-1-r=\dfrac{13}{12}\)

Multiplying by \(-12r\): \(12+12r+12r^2=-13r \Rightarrow 12r^2+25r+12=0\)

\(r=\dfrac{-25\pm\sqrt{625-576}}{24}=\dfrac{-25\pm7}{24}\), giving \(r=-\dfrac34\) or \(r=-\dfrac43\)

Taking \(r=-\dfrac34\): terms are \(\dfrac{-1}{-3/4},\ -1,\ (-1)\left(-\dfrac34\right) = \dfrac43,\ -1,\ \dfrac34\)

Check: sum \(=\dfrac43-1+\dfrac34=\dfrac{16}{12}-\dfrac{12}{12}+\dfrac{9}{12}=\dfrac{13}{12}\) ✓; product \(=\dfrac43\times(-1)\times\dfrac34=-1\) ✓

Common ratio = \(-\dfrac{3}{4}\) (or equivalently \(-\dfrac{4}{3}\), same terms in reverse); the terms are \(\dfrac{4}{3}\), −1, \(\dfrac{3}{4}\).

12*If the 4th, 10th and 16th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.

Let the GP have first term a and common ratio r.

\(x=t_4=ar^3\), \(y=t_{10}=ar^9\), \(z=t_{16}=ar^{15}\)

To show x, y, z are in GP, we need \(y^2=xz\):

\(y^2=(ar^9)^2=a^2r^{18}\)

\(xz=(ar^3)(ar^{15})=a^2r^{18}\)

Since \(y^2=xz\), x, y, z are in GP (with common ratio \(r^6\)). Hence proved.

y² = xz = a²r¹⁸, so x, y, z are in GP with common ratio r⁶.

13*The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP.

Let the terms be \(\dfrac{a}{r}, a, ar\).

Sum: \(a\left(\dfrac1r+1+r\right)=26\) …(1)

Sum of squares: \(a^2\left(\dfrac{1}{r^2}+1+r^2\right)=364\) …(2)

Let \(P=\dfrac1r+1+r\). Then \(P^2 = \left(\dfrac{1}{r^2}+1+r^2\right)+2P\), so from (1) and (2): \(\dfrac{364}{a^2}=\left(\dfrac{26}{a}\right)^2-2\left(\dfrac{26}{a}\right)\)

Multiplying by \(a^2\): \(364=676-52a \Rightarrow 52a=312 \Rightarrow a=6\)

Then \(P=\dfrac{26}{6}=\dfrac{13}{3}\), so \(\dfrac1r+r=\dfrac{13}{3}-1=\dfrac{10}{3}\)

\(3+3r^2=10r \Rightarrow 3r^2-10r+3=0 \Rightarrow r=\dfrac{10\pm\sqrt{100-36}}{6}=\dfrac{10\pm8}{6}\)

\(r=3\) or \(r=\dfrac13\). Taking \(r=3\): terms are \(\dfrac{6}{3},6,6(3) = 2,6,18\)

Check: sum \(=2+6+18=26\) ✓; sum of squares \(=4+36+324=364\) ✓

The terms of the GP are 2, 6, 18.

14*Suppose P₁ = 1, P₂ = 2 and for n > 2, Pₙ = P₁ + P₂ + ⋯ + Pₙ₋₁ + 1. Find the values of P₁, P₂, ..., P₈. Can you find a simpler recursive formula for Pₙ? Can you give an explicit formula?

\(P_1=1,\ P_2=2\)

\(P_3=P_1+P_2+1=1+2+1=4\)

\(P_4=P_1+P_2+P_3+1=1+2+4+1=8\)

\(P_5=1+2+4+8+1=16\)

\(P_6=1+2+4+8+16+1=32\)

\(P_7=1+2+4+8+16+32+1=64\)

\(P_8=1+2+4+8+16+32+64+1=128\)

The sequence 1, 2, 4, 8, 16, 32, 64, 128 is a GP with \(a=1, r=2\). Since \(P_n=P_1+\cdots+P_{n-1}+1\) and \(P_{n-1}=P_1+\cdots+P_{n-2}+1\), subtracting gives \(P_n-P_{n-1}=P_{n-1}\), i.e. \(P_n=2P_{n-1}\) — this holds for all \(n\ge2\).

P₁,...,P₈ = 1, 2, 4, 8, 16, 32, 64, 128. Simpler recursive formula: P₁ = 1, Pₙ = 2Pₙ₋₁ for n ≥ 2. Explicit formula: Pₙ = 2ⁿ⁻¹.

15*Suppose W₁ = 1, W₂ = 2 and for n > 2, Wₙ = W₁ + W₂ + ⋯ + Wₙ₋₂ + 2. Find the values of W₁, W₂, ..., W₈. Do you recognise this sequence?

\(W_1=1,\ W_2=2\)

\(W_3=W_1+2=1+2=3\)

\(W_4=W_1+W_2+2=1+2+2=5\)

\(W_5=W_1+W_2+W_3+2=1+2+3+2=8\)

\(W_6=1+2+3+5+2=13\)

\(W_7=1+2+3+5+8+2=21\)

\(W_8=1+2+3+5+8+13+2=34\)

W₁,...,W₈ = 1, 2, 3, 5, 8, 13, 21, 34 — this is the Virahānka–Fibonacci sequence!

Frequently Asked Questions

An explicit formula gives a term's value directly from its position number n (e.g. tₙ = 2n − 1), so you can find any term without knowing the others. A recursive formula gives a term's value using the value(s) of the previous term(s) (e.g. t₁ = 1, tₙ = tₙ₋₁ + 3), so you need to know earlier terms first.

tₙ = a + (n − 1)d, where a is the first term and d is the common difference (the constant amount added to get from one term to the next).

tₙ = ar ⁿ⁻¹, where a is the first term and r is the common ratio (the constant factor you multiply by to get from one term to the next).

Sₙ = n(n + 1) / 2. This is derived by writing the sum forwards and backwards and adding the two, and it is also the formula for the nth triangular number.

It's the sequence 1, 2, 3, 5, 8, 13, 21, 34, ... where each term is the sum of the previous two. It was first written down by the Indian scholar Virahānka in the 7th century CE while studying Prakrit poetic meter, centuries before it was studied in Europe by Fibonacci.

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Chapter 8Predicting What Comes Next: Sequences and Progressions