Predicting What Comes Next: Exploring Sequences and Progressions

1, 2, 3, 4, 5, 6, … (Natural Numbers)
1, 3, 5, 7, 9, 11, … (Odd Numbers)
1, 3, 6, 10, 15, 21, … (Triangular Numbers)
1, 4, 9, 16, 25, 36, … (Square Numbers)

Can you describe the pattern in each of the above sequences? Can you predict the next few numbers in these sequences?

Answer

Natural Numbers : 1, 2, 3, 4, 5, 6, …

Each term is 1 more than the previous term. So, the next few numbers are 6 + 1 = 7, 7 + 1 = 8, 8 + 1 = 9, 9 + 1 = 10, …

i.e. 7, 8, 9, 10, …

Odd Numbers : 1, 3, 5, 7, 9, 11, …

Each term is 2 more than the previous term. So, the next few numbers are 11 + 2 = 13, 13 + 2 = 15, 15 + 2 = 17, 17 + 2 = 19, …

i.e. 13, 15, 17, 19, …

Triangular Numbers : 1, 3, 6, 10, 15, 21, …

Each term is the sum of the natural numbers up to that term (1 = 1, 3 = 1 + 2, 6 = 1 + 2 + 3, 10 = 1 + 2 + 3 + 4, …). The differences between consecutive terms are 2, 3, 4, 5, 6, …, so the next differences are 7, 8, 9, 10.

The next few numbers are 21 + 7 = 28, 28 + 8 = 36, 36 + 9 = 45, 45 + 10 = 55, …

i.e. 28, 36, 45, 55, …

Square Numbers : 1, 4, 9, 16, 25, 36, …

Each term is the square of its position number (1 = 1², 4 = 2², 9 = 3², …). Also, each term is the sum of consecutive odd numbers (1 = 1, 4 = 1 + 3, 9 = 1 + 3 + 5, …), so the differences between consecutive terms are 3, 5, 7, 9, 11, … and the next differences are 13, 15, 17, 19.

The next few numbers are 36 + 13 = 49, 49 + 15 = 64, 64 + 17 = 81, 81 + 19 = 100, …

i.e. 49, 64, 81, 100, …

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

Answer

Yes, there are many other kinds of sequences. Five different types of sequences along with their properties are given below.

1. Even Numbers: 2, 4, 6, 8, 10, …
Each term is 2 more than the previous term, so the difference between consecutive terms is constant (= 2). The n^th term is t_n = 2n.

2. Multiples of 5: 5, 10, 15, 20, 25, …
Each term is obtained by adding 5 to the previous term, so the difference between consecutive terms is constant (= 5). The n^th term is t_n = 5n.

3. Powers of 2: 1, 2, 4, 8, 16, 32, …
Each term is twice the previous term, so the ratio of consecutive terms is constant (= 2). The n^th term is t_n = 2^{n - 1}.

4. Unit Fractions: $1, \dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}, \dfrac{1}{5}, …$
The terms are fractions occurring in decreasing order, getting closer and closer to 0. The n^th term is t_n = $\dfrac{1}{n}$.

5. Virahānka–Fibonacci Sequence: 1, 2, 3, 5, 8, 13, 21, …
Each term from the third term onwards is the sum of the two preceding terms, i.e. t_n = t_{n - 1} + t_{n - 2} for n ≥ 3.

Why is it useful to have an explicit formula for the n^th term of a sequence?

Answer

An explicit formula uses the position number n to calculate the value of a term directly. It is useful because:

1. We can find any term of the sequence (such as the 20th, 53rd, 300th or any other term) directly by just substituting the value of n in the formula, without having to know the values of the previous terms.

2. We can use it to check whether a given number is a term of the sequence and, if it is, to find the position it occupies in the sequence.

Using the explicit rule u_n = 2n – 1, find the 53^rd, 108^th and 1170^th terms of the odd-number sequence.

Answer

Using u_n = 2n - 1,

u₅₃ = 2(53) - 1 = 106 - 1 = 105,

u₁₀₈ = 2(108) - 1 = 216 - 1 = 215,

u₁₁₇₀ = 2(1170) - 1 = 2340 - 1 = 2339.

Hence, the 53^rd, 108^th and 1170^th terms are 105, 215 and 2339 respectively.

Can you find the rule describing the n^th term of the sequence of square numbers?

Answer

The sequence of square numbers is 1, 4, 9, 16, 25, 36, …

Each term is the square of its position number:
1 = 1², $\quad$ 4 = 2², $\quad$ 9 = 3², $\quad$ 16 = 4², $\quad$ 25 = 5², $\quad$ 36 = 6², …

So the term in the nth position is the square of n.

Hence, the n^th term of the sequence of square numbers is t_n = n².

Consider the expression t_n = 3n – 7.

(i) Find its first, second, third, 12^th, 18^th and 50^th terms.

(ii) Which term of the sequence is 332?

(iii) Is 557 a term of this sequence? Why or why not?

Answer

(i) Using t_n = 3n - 7,

t₁ = 3(1) - 7 = -4,
t₂ = 3(2) - 7 = -1,
t₃ = 3(3) - 7 = 2,
t₁₂ = 3(12) - 7 = 29,
t₁₈ = 3(18) - 7 = 47,
t₅₀ = 3(50) - 7 = 143.

(ii) Let 332 be the n^th term.

⇒ 3n - 7 = 332
⇒ 3n = 339
⇒ n = 113.

Thus, 332 is the 113^th term.

(iii) Let 557 be the n^th term.

⇒ 3n - 7 = 557
⇒ 3n = 564
⇒ n = 188.

Since 188 is a natural number, 557 is a term of the sequence; it is the 188^th term.

Hence, the required terms are -4, -1, 2, 29, 47 and 143; 332 is the 113^th term; and 557 is the 188^th term.

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

Answer

(i) Given, t_n = 3n - 4.

Putting n = 1, 2, 3, 4, 5 we get,

t₁ = 3(1) - 4 = 3 - 4 = -1,
t₂ = 3(2) - 4 = 6 - 4 = 2,
t₃ = 3(3) - 4 = 9 - 4 = 5,
t₄ = 3(4) - 4 = 12 - 4 = 8,
t₅ = 3(5) - 4 = 15 - 4 = 11.

Hence, the first five terms of the sequence are -1, 2, 5, 8, 11.

(ii) Given, t_n = 2 - 5n.

Putting n = 1, 2, 3, 4, 5 we get,

t₁ = 2 - 5(1) = 2 - 5 = -3,
t₂ = 2 - 5(2) = 2 - 10 = -8,
t₃ = 2 - 5(3) = 2 - 15 = -13,
t₄ = 2 - 5(4) = 2 - 20 = -18,
t₅ = 2 - 5(5) = 2 - 25 = -23.

Hence, the first five terms of the sequence are -3, -8, -13, -18, -23.

(iii) Given, t_n = n² - 2n + 3.

Putting n = 1, 2, 3, 4, 5 we get,

t₁ = (1)² - 2(1) + 3 = 1 - 2 + 3 = 2,
t₂ = (2)² - 2(2) + 3 = 4 - 4 + 3 = 3,
t₃ = (3)² - 2(3) + 3 = 9 - 6 + 3 = 6,
t₄ = (4)² - 2(4) + 3 = 16 - 8 + 3 = 11,
t₅ = (5)² - 2(5) + 3 = 25 - 10 + 3 = 18.

Hence, the first five terms of the sequence are 2, 3, 6, 11, 18.

Find the 10^th and 15^th terms of the sequence t_n = 5n – 3 for n ≥ 1.

Answer

Given, t_n = 5n - 3.

Putting n = 10, for the 10th term we get,

t₁₀ = 5(10) - 3 = 50 - 3 = 47.

Putting n = 15, for the 15th term we get,

t₁₅ = 5(15) - 3 = 75 - 3 = 72.

Hence, the 10th term is 47 and the 15th term is 72.

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

Answer

Given, t_n = 5n - 3.

A number is a term of the sequence only if the value of n obtained on solving t_n = (that number) is a natural number.

For 97:

Let 97 be the n^th term, i.e. t_n = 97.

⇒ 5n - 3 = 97

⇒ 5n = 97 + 3

⇒ 5n = 100

⇒ n = $\dfrac{100}{5}$

⇒ n = 20, which is a natural number.

So, 97 is a term of the sequence (the 20th term).

For 172:

Let 172 be the n^th term, i.e. t_n = 172.

⇒ 5n - 3 = 172

⇒ 5n = 172 + 3

⇒ 5n = 175

⇒ n = $\dfrac{175}{5}$

⇒ n = 35, which is a natural number.

So, 172 is a term of the sequence (the 35th term).

Hence, both 97 and 172 are terms of the sequence, being the 20th and 35th terms respectively.

Which term of the sequence t_n = 5n – 3 for n ≥ 1 is 607?

Answer

Given, t_n = 5n - 3.

Let 607 be the n^th term of the sequence, i.e. t_n = 607.

⇒ 5n - 3 = 607

⇒ 5n = 607 + 3

⇒ 5n = 610

⇒ n = $\dfrac{610}{5}$

⇒ n = 122.

Hence, 607 is the 122^nd term of the sequence.

A sequence is given by the recursive rule t₁ = – 5, t_n+1 = t_n + 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?

Answer

Given, t₁ = -5 and t_n+1 = t_n + 3 for n ≥ 1.

Putting n = 1, 2, 3, 4 successively we get,

t₂ = t₁ + 3 = -5 + 3 = -2,
t₃ = t₂ + 3 = -2 + 3 = 1,
t₄ = t₃ + 3 = 1 + 3 = 4,
t₅ = t₄ + 3 = 4 + 3 = 7.

∴ The first five terms of the sequence are -5, -2, 1, 4, 7.

Since each term is obtained by adding 3 to the previous term, the sequence is an AP with first term a = -5 and common difference d = 3.

∴ n^th term = t_n = a + (n - 1)d = -5 + (n - 1)(3) = -5 + 3n - 3 = 3n - 8.

Let 52 be the n^th term, i.e. t_n = 52.

⇒ 3n - 8 = 52

⇒ 3n = 52 + 8

⇒ 3n = 60

⇒ n = $\dfrac{60}{3}$

⇒ n = 20.

Hence, the first five terms of the sequence are -5, -2, 1, 4, 7 and 52 is a term of the sequence, being its 20th term.

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

Answer

Given, T₁ = 1, T₂ = 2, T₃ = 4 and T_n = T_n-1 + T_n-2 + T_n-3 for n ≥ 4.

Putting n = 4, 5, 6, 7, 8 successively we get,

T₄ = T₃ + T₂ + T₁ = 4 + 2 + 1 = 7,
T₅ = T₄ + T₃ + T₂ = 7 + 4 + 2 = 13,
T₆ = T₅ + T₄ + T₃ = 13 + 7 + 4 = 24,
T₇ = T₆ + T₅ + T₄ = 24 + 13 + 7 = 44,
T₈ = T₇ + T₆ + T₅ = 44 + 24 + 13 = 81.

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

Can you predict the number of squares in Stages 5 and 6 of the sequence? In Stages 10, 11 and 12? In Stage 20? At any stage?

Answer

The number of tiny squares at successive stages forms the sequence 1, 5, 9, 13, …

At each stage, 4 squares get added to the corners of the pattern in the earlier stage, so the difference between consecutive terms is the fixed number 4. Hence the sequence is an AP with first term a = 1 and common difference d = 4.

The number of squares at the nth stage is given by:

t_n = a + (n - 1)d = 1 + (n - 1) × 4 = 4n - 3.

Putting the required values of n, we get:

Stage 5: t₅ = 4 × 5 - 3 = 20 - 3 = 17,
Stage 6: t₆ = 4 × 6 - 3 = 24 - 3 = 21,
Stage 10: t₁₀ = 4 × 10 - 3 = 40 - 3 = 37,
Stage 11: t₁₁ = 4 × 11 - 3 = 44 - 3 = 41,
Stage 12: t₁₂ = 4 × 12 - 3 = 48 - 3 = 45,
Stage 20: t₂₀ = 4 × 20 - 3 = 80 - 3 = 77.

Hence, the number of squares in Stage 5 = 17, Stage 6 = 21; Stage 10 = 37, Stage 11 = 41, Stage 12 = 45; Stage 20 = 77; and 4n - 3 at any stage n.

Consider all the sequences discussed so far in the chapter. Which ones are arithmetic progressions and which ones are not? Justify your answer.

Answer

A sequence is an AP when the difference between every pair of consecutive terms is constant.

The following sequences are APs:

• Natural numbers 1, 2, 3, 4, …, with common difference 1.
• Odd numbers 1, 3, 5, 7, …, with common difference 2.
• The sequence 1, 4, 7, 10, …, with common difference 3.
• The sequence -7, -3, 1, 5, …, with common difference 4.
• Any sequence given by a linear rule such as s_n = 5n - 2 or t_n = 3n - 7.

The following sequences are not APs:

• Triangular numbers 1, 3, 6, 10, …, because the consecutive differences are 2, 3, 4, … .
• Square numbers 1, 4, 9, 16, …, because the consecutive differences are 3, 5, 7, … .
• Unit fractions 1, $\dfrac{1}{2}$, $\dfrac{1}{3}$, $\dfrac{1}{4}$, …, because their consecutive differences are not constant.
• Prime numbers 2, 3, 5, 7, 11, …, because their consecutive differences are not constant.
• The Virahānka–Fibonacci sequence 1, 2, 3, 5, 8, …, because its consecutive differences are not constant.

Thus, only the sequences having a fixed difference between consecutive terms are arithmetic progressions.

Verify that the following sequences are arithmetic progressions and write their n^th terms. What do you observe when you plot the ordered pairs obtained from them?

(i) 2, 5, 8, 11, …

(ii) -5, -1, 3, 7, …

Answer

(i) The consecutive differences are

5 - 2 = 3, 8 - 5 = 3 and 11 - 8 = 3.

Therefore, the sequence is an AP with a = 2 and d = 3.

$t_n = a + (n - 1)d = 2 + 3(n - 1) = 3n - 1.$

(ii) The consecutive differences are

-1 - (-5) = 4, 3 - (-1) = 4 and 7 - 3 = 4.

Therefore, the sequence is an AP with a = -5 and d = 4.

$t_n = a + (n - 1)d = -5 + 4(n - 1) = 4n - 9.$

In each case, the rule for t_n is linear in n. Therefore, the ordered pairs (n, t_n) lie on a straight line.

Hence, the n^th terms are t_n = 3n - 1 and t_n = 4n - 9 respectively, and the corresponding plotted points lie on straight lines.

Using the formula t_n = a + (n - 1)d, find the n^th term of the following arithmetic progressions.

(i) $\dfrac{1}{2}, \dfrac{5}{2}, \dfrac{9}{2}, \dfrac{13}{2}, …$

(ii) 1.5, 3.5, 5.5, 7.5, …

Answer

(i) Here, a = $\dfrac{1}{2}$ and d = 2.

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

(ii) Here, a = 1.5 and d = 2.

$t_n = 1.5 + (n - 1)2 = 2n - 0.5.$

Hence, the n^th terms are $t_n = \dfrac{4n - 3}{2}$ and $t_n = 2n - 0.5$ respectively.

Find recursive rules for the APs in the previous two exercises.

Answer

The recursive rule of an AP is t₁ = a and t_n = t_{n - 1} + d for n ≥ 2.

Therefore:

(i) For 2, 5, 8, 11, … :

t₁ = 2, t_n = t_{n - 1} + 3 for n ≥ 2.

(ii) For -5, -1, 3, 7, … :

t₁ = -5, t_n = t_{n - 1} + 4 for n ≥ 2.

(iii) For $\dfrac{1}{2}, \dfrac{5}{2}, \dfrac{9}{2}, …$:

t₁ = $\dfrac{1}{2}$, t_n = t_{n - 1} + 2 for n ≥ 2.

(iv) For 1.5, 3.5, 5.5, … :

t₁ = 1.5, t_n = t_{n - 1} + 2 for n ≥ 2.

To find the sum of the first ten natural numbers without actually adding all of them, we can use the below method:

S = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
S = 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1

Each pair of corresponding numbers, 1 + 10, 2 + 9, 3 + 8, ... , 10 + 1, in the two equations add up to 11.

2S = 11 + 11 + 11 + … + 11 (that is, 11 added 10 times).

This leads to 2S = 110 or S = 55.

Can the same approach be used to find the sum of 1 + 2 + 3 + … + 100?

Answer

Yes, the same approach can be used.

Let S denote the sum of the first hundred natural numbers. Thus,

S = 1 + 2 + 3 + … + 99 + 100.

We can also write it in the reverse order as

S = 100 + 99 + 98 + … + 2 + 1.

Placing these two equations one below the other,

S = 1 + 2 + 3 + … + 99 + 100
S = 100 + 99 + 98 + … + 2 + 1

each pair of corresponding numbers, 1 + 100, 2 + 99, 3 + 98, … , 100 + 1, adds up to 101.

Adding the two equations for S, we get

2S = 101 + 101 + 101 + … + 101 (that is, 101 added 100 times).

⇒ 2S = 101 × 100

⇒ 2S = 10100

⇒ S = $\dfrac{10100}{2}$

⇒ S = 5050.

Hence, the sum 1 + 2 + 3 + … + 100 = 5050.

Can you use the formula S_n = $\dfrac{n(n + 1)}{2}$ to find S₂₀, S₅₀ or S₁₀₀₀?

Answer

Yes, we can substitute the required value of n in the formula S_n = $\dfrac{n(n + 1)}{2}$.

Putting n = 20,

S₂₀ = $\dfrac{20 \times (20 + 1)}{2} = \dfrac{20 \times 21}{2} = \dfrac{420}{2} = 210.$

Putting n = 50,

S₅₀ = $\dfrac{50 \times (50 + 1)}{2} = \dfrac{50 \times 51}{2} = \dfrac{2550}{2} = 1275.$

Putting n = 1000,

S₁₀₀₀ = $\dfrac{1000 \times (1000 + 1)}{2} = \dfrac{1000 \times 1001}{2} = \dfrac{1001000}{2} = 500500.$

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

Let us revisit the sequence t_n of triangular numbers 1, 3, 6, 10, 15, … shown in Fig. 8.1. Note that the n^th term of this sequence is the sum of the first n natural numbers. Thus t_n = $\dfrac{n(n+1)}{2}$.

Can you use this to find the 10^th, 17^th and 80^th triangular numbers?

Answer

Yes, we can substitute the required value of n in the formula t_n = $\dfrac{n(n + 1)}{2}$.

Putting n = 10, the 10^th triangular number is

t₁₀ = $\dfrac{10 \times (10 + 1)}{2} = \dfrac{10 \times 11}{2} = \dfrac{110}{2} = 55.$

Putting n = 17, the 17^th triangular number is

t₁₇ = $\dfrac{17 \times (17 + 1)}{2} = \dfrac{17 \times 18}{2} = \dfrac{306}{2} = 153.$

Putting n = 80, the 80^th triangular number is

t₈₀ = $\dfrac{80 \times (80 + 1)}{2} = \dfrac{80 \times 81}{2} = \dfrac{6480}{2} = 3240.$

Hence, the 10^th, 17^th and 80^th triangular numbers are 55, 153 and 3240 respectively.

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

Answer

The given AP is 3, 8, 13, 18, ….

Here, first term a = 3

common difference d = 8 - 3 = 5.

The n^th term of an AP is given by:

t_n = a + (n - 1)d.

∴ t_n = 3 + (n - 1)5

= 3 + 5n - 5

= 5n - 2.

Putting n = 10, we get

t₁₀ = 5 × 10 - 2

= 50 - 2

= 48.

Putting n = 26, we get

t₂₆ = 5 × 26 - 2

= 130 - 2

= 128.

Hence, the 10^th term is 48 and the 26^th term is 128.

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

Answer

The given AP is 21, 18, 15, ….

Here, first term a = 21

common difference d = 18 - 21 = -3.

The n^th term of an AP is given by:

t_n = a + (n - 1)d.

∴ t_n = 21 + (n - 1)(-3)

= 21 - 3n + 3

= 24 - 3n.

Let -81 be the n^th term of the AP, i.e. t_n = -81.

⇒ 24 - 3n = -81

⇒ 3n = 24 + 81

⇒ 3n = 105

⇒ n = $\dfrac{105}{3}$

⇒ n = 35.

So, -81 is the 35^th term of the AP.

Now, let us check whether 0 is a term of the AP, i.e. t_n = 0.

⇒ 24 - 3n = 0

⇒ 3n = 24

⇒ n = $\dfrac{24}{3}$

⇒ n = 8.

Since n = 8 is a natural number, 0 is the 8^th term of the AP.

Hence, -81 is the 35^th term of the AP, and 0 is the 8^th term of the AP.

Find the n^th term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.

Answer

The given AP is 11, 8, 5, 2, ….

Here, first term a = 11

common difference d = 8 - 11 = -3.

The n^th term of an AP is given by:

t_n = a + (n - 1)d.

∴ t_n = 11 + (n - 1)(-3)

= 11 - 3n + 3

= 14 - 3n.

Since every term is 3 less than its previous term, the recursive rule is:

t₁ = 11, t_n = t_{n - 1} - 3 for n ≥ 2.

Hence, the n^th term is t_n = 14 - 3n and the recursive rule is t₁ = 11, t_n = t_{n - 1} - 3 for n ≥ 2.

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

(Hint: If 'a' is the first term and 'd' the common difference, then we arrive at the equations a + 2d = 12 and a + 49d = 106. Solve this pair of linear equations for 'a' and 'd'.)

Answer

Let the first term be a and the common difference be d.

The n^th term of an AP is given by:

t_n = a + (n - 1)d.

Given, the 3^rd term is 12.

⇒ t₃ = a + (3 - 1)d = 12

⇒ a + 2d = 12 $\quad$ (Eq 1)

Since the AP has 50 terms, the last term is the 50^th term, which is 106.

⇒ t₅₀ = a + (50 - 1)d = 106

⇒ a + 49d = 106 $\quad$ (Eq 2)

Subtracting Eq 1 from Eq 2,

(a + 49d) - (a + 2d) = 106 - 12

⇒ 47d = 94

⇒ d = 2.

Putting d = 2 in Eq 1,

⇒ a + 2 × 2 = 12

⇒ a + 4 = 12

⇒ a = 8.

Now, the 29^th term is

t₂₉ = a + (29 - 1)d

= 8 + 28 × 2

= 8 + 56

= 64.

Hence, the 29^th term of the AP is 64.

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

Answer

The 2-digit numbers which are divisible by 3 are:

12, 15, 18, 21, …, 99.

These numbers form an AP with:

first term a = 12,

common difference d = 15 - 12 = 3

last term = l = 99.

Let 99 be the n^th term, i.e. t_n = 99.

Using the formula, t_n = a + (n - 1)d

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

$\Rightarrow$ 99 = 12 + 3n - 3

$\Rightarrow$ 99 = 9 + 3n

$\Rightarrow$ 3n = 90

$\Rightarrow$ n = 30.

So, there are 30 such numbers.

Let S be the sum of these numbers. Writing S in two ways:

$S \phantom{00} = 12 + 15 + 18 + \cdots + 96 + 99$

$S \phantom{00} = 99 + 96 + 93 + \cdots + 15 + 12$

Adding both rows term by term, each pair adds up to 12 + 99 = 111, and there are 30 such pairs:

$2S = 111 + 111 + 111 + \cdots \text{ (30 times)}$

$2S = 30 \times 111$

$2S = 3330$

$S = \dfrac{3330}{2} = 1665.$

Hence, there are 30 two-digit numbers divisible by 3 and their sum is 1665.

Harish 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?

Answer

Harish's annual salary in successive years is

₹5,00,000, ₹5,20,000, ₹5,40,000, …

This forms an AP with first term a = 5,00,000

common difference d = 20,000.

Let his income reach ₹7,00,000 in the nth year, i.e. t_n = 7,00,000.

By formula, t_n = a + (n - 1)d

⇒ 7,00,000 = 5,00,000 + (n - 1) × 20,000

⇒ 7,00,000 - 5,00,000 = (n - 1) × 20,000

⇒ 2,00,000 = (n - 1) × 20,000

⇒ n - 1 = $\dfrac{2,00,000}{20,000}$

⇒ n - 1 = 10

⇒ n = 11.

So his income reaches ₹7,00,000 in the 11^th year, that is, after 10 years of increments.

Hence, after 10 years (in the 11^th year) Harish's income reached ₹7,00,000.

A 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?

Answer

The number of marbles in successive rows is 1, 2, 3, 4, ……, 25.

So, the total number of marbles used is the sum of the first 25 natural numbers.

Total marbles = 1 + 2 + 3 + … + 25 = S₂₅.

By formula, S_n = $\dfrac{n(n + 1)}{2}$

Putting n = 25,

$S_{25} = \dfrac{25 \times (25 + 1)}{2} \\[1em] = \dfrac{25 \times 26}{2} \\[1em] = \dfrac{650}{2} \\[1em] = 325.$

Hence, the child uses 325 marbles in all.

Can you predict the number of squares in Stages 5 and 6 of the pattern? In Stages 10, 11 and 12? In Stage 20? At any stage? How is this different from the growing pattern in Fig. 8.3?

Answer

The number of green squares in the four stages of the pattern (Fig. 8.6) gives the list of numbers 3, 6, 12, 24, …

Here, $\dfrac{6}{3} = \dfrac{12}{6} = \dfrac{24}{12} = 2$, a fixed number.

So, the given list of numbers forms a GP with first term a = 3 and common ratio r = 2.

By formula, the number of squares at the nth stage = t_n = ar^{n - 1} = 3 × 2^{n - 1}.

Putting n = 5 and n = 6, the number of squares in Stages 5 and 6 are:

t₅ = 3 × 2^{5 - 1} = 3 × 2⁴ = 3 × 16 = 48,
t₆ = 3 × 2^{6 - 1} = 3 × 2⁵ = 3 × 32 = 96.

Putting n = 10, 11 and 12, the number of squares in Stages 10, 11 and 12 are:

t₁₀ = 3 × 2^{10 - 1} = 3 × 2⁹ = 3 × 512 = 1536,
t₁₁ = 3 × 2^{11 - 1} = 3 × 2¹⁰ = 3 × 1024 = 3072,
t₁₂ = 3 × 2^{12 - 1} = 3 × 2¹¹ = 3 × 2048 = 6144.

Putting n = 20, the number of squares in Stage 20 is:

t₂₀ = 3 × 2^{20 - 1} = 3 × 2¹⁹ = 3 × 524288 = 1572864.

At any stage, the number of squares = t_n = 3 × 2^{n - 1}.

Hence, the number of squares in Stage 5 = 48 and Stage 6 = 96; Stage 10 = 1536, Stage 11 = 3072 and Stage 12 = 6144; Stage 20 = 1572864; and 3 × 2^{n - 1} at any stage n.

How it is different from Fig. 8.3: In Fig. 8.3, the number of squares (1, 5, 9, 13, …) increases by adding a fixed number 4 at each stage, so the difference between consecutive terms is constant — it forms an AP In Fig. 8.6, the number of squares (3, 6, 12, 24, …) is doubled at each stage, so the ratio between consecutive terms is constant — it forms a GP Thus, the pattern in Fig. 8.3 grows by a fixed amount, while the pattern in Fig. 8.6 grows by a fixed multiple and therefore increases much more rapidly.

Check whether the following sequences are geometric progressions and find their n^th terms.

(i) 2, 10, 50, 250, …

(ii) $4, \dfrac{8}{3}, \dfrac{16}{9}, \dfrac{32}{27}, …$

(iii) $3, -\dfrac{3}{2}, \dfrac{3}{4}, -\dfrac{3}{8}, …$

Answer

(i)

$\dfrac{10}{2} = \dfrac{50}{10} = \dfrac{250}{50} = 5.$

Thus, the sequence is a GP with a = 2 and r = 5.

$t_n = ar^{n - 1} = 2 \times 5^{n - 1}.$

(ii)

$\dfrac{\frac{8}{3}}{4} = \dfrac{\frac{16}{9}}{\frac{8}{3}} = \dfrac{\frac{32}{27}}{\frac{16}{9}} = \dfrac{2}{3}.$

Thus, the sequence is a GP with a = 4 and r = $\dfrac{2}{3}$.

$t_n = 4\left(\dfrac{2}{3}\right)^{n - 1}.$

(iii)

$\dfrac{-\frac{3}{2}}{3} = \dfrac{\frac{3}{4}}{-\frac{3}{2}} = \dfrac{-\frac{3}{8}}{\frac{3}{4}} = -\dfrac{1}{2}.$

Thus, the sequence is a GP with a = 3 and r = $-\dfrac{1}{2}$.

$t_n = 3\left(-\dfrac{1}{2}\right)^{n - 1}.$

Hence, all three sequences are GPs. Their n^th terms are $2 \times 5^{n - 1}$, $4\left(\dfrac{2}{3}\right)^{n - 1}$ and $3\left(-\dfrac{1}{2}\right)^{n - 1}$ respectively.

Can you find a recursive rule for the formula t_n = 3 × 10^{n - 1} that generates the geometric progression 3, 30, 300, 3000, …?

Answer

The first term is 3, and every term after the first is 10 times the preceding term.

Therefore, the recursive rule is

t₁ = 3, t_n = 10t_{n - 1} for n ≥ 2.

Hence, the required recursive rule is t₁ = 3 and t_n = 10t_{n - 1} for n ≥ 2.

Observe the Sierpiński triangle and try to answer the following questions

(a) How many black triangles are there in Stages 0 to 3 of Fig. 8.7?

(b) Can you predict the number of black triangles at Stages 4 and 5?

(c) Can you find a rule for the number of black triangles at the n^th stage?

(d) Suppose the area of the triangle (that is, the black region) in Stage 0 is 1 square unit. What is the area of the black region in Stages 1, 2 and 3? What will be the area of the black region in Stages 4 and 5? Find a rule for the area of the black region at the n^th stage. What happens to this area as n, the number of stages, goes on increasing?

Answer

(a) At every stage, each black triangle is replaced by three smaller black triangles at the next stage. So the number of black triangles becomes three times the number at the previous stage.

Stage 0: 1 black triangle,

Stage 1: 1 × 3 = 3 black triangles,

Stage 2: 3 × 3 = 9 black triangles,

Stage 3: 9 × 3 = 27 black triangles.

Hence, the numbers of black triangles in Stages 0 to 3 are 1, 3, 9 and 27.

(b) Continuing the same way (multiplying by 3 each time),

Stage 4: 27 × 3 = 81 black triangles,

Stage 5: 81 × 3 = 243 black triangles.

Hence, the numbers of black triangles at Stages 4 and 5 are 81 and 243.

(c) The number of black triangles forms the list 1, 3, 9, 27, 81, 243, …

Here, $\dfrac{3}{1} = \dfrac{9}{3} = \dfrac{27}{9} = 3$, a fixed number, so it is a GP

All these terms are powers of 3:

1 = 3⁰, 3 = 3¹, 9 = 3², 27 = 3³, …

The exponent of 3 matches the stage number.

Hence, the number of black triangles at the n^th stage is t_n = 3ⁿ.

(d) In Stage 1, the triangle at Stage 0 is divided into 4 equal parts and the central part is removed. So $\dfrac{3}{4}$ of the area remains. The same process is repeated at every stage, so the area is multiplied by $\dfrac{3}{4}$ each time.

Area at Stage 1 = $1 \times \dfrac{3}{4} = \dfrac{3}{4}$ square units,

Area at Stage 2 = $\dfrac{3}{4} \times \dfrac{3}{4} = \Big(\dfrac{3}{4}\Big)^2 = \dfrac{9}{16}$ square units,

Area at Stage 3 = $\dfrac{9}{16} \times \dfrac{3}{4} = \Big(\dfrac{3}{4}\Big)^3 = \dfrac{27}{64}$ square units.

For Stages 4 and 5,

Area at Stage 4 = $\Big(\dfrac{3}{4}\Big)^4 = \dfrac{81}{256}$ square units,

Area at Stage 5 = $\Big(\dfrac{3}{4}\Big)^5 = \dfrac{243}{1024}$ square units.

The areas form a GP with first term 1 and common ratio $\dfrac{3}{4}$.

Hence, the area of the black region at the n^th stage is s_n = $\Big(\dfrac{3}{4}\Big)^n$.

As n goes on increasing, the common ratio $\dfrac{3}{4}$ is less than 1, so each term is smaller than the previous one.

Thus, the area of the black region keeps decreasing and gets closer and closer to 0.

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

Answer

Given, a₈ = 192 and r = 2.

By formula, a_n = ar^{n - 1}.

⇒ a₈ = a(2)^{(8 - 1)}

⇒ 192 = a(2)⁷

⇒ a = $\dfrac{192}{2^7} = \dfrac{192}{128} = \dfrac{3}{2}.$

12th term of the GP is a₁₂,

$\Rightarrow a_{12} = \dfrac{3}{2}(2)^{12 - 1} \\[1em] \Rightarrow a_{12} = \dfrac{3}{2} \times 2^{11} \\[1em] \Rightarrow a_{12} = 3 \times 2^{10} \\[1em] \Rightarrow a_{12} = 3 \times 1024 \\[1em] \Rightarrow a_{12} = 3072.$

Hence, the 12^th term of the GP is 3072.

Find the 10^th and n^th terms of the GP: 5, 25, 125, … .

Answer

The given list of numbers 5, 25, 125, … is a GP

first term a = 5

common ratio r = $\dfrac{25}{5}$ = 5.

By formula, a_n = ar^{n - 1}

a₁₀ = 5(5)^{10 - 1} = 5(5)⁹ = 5¹⁰.

a_n = 5(5)^{n - 1} = 5 x 5ⁿ x 5^-1 = 5ⁿ.

Hence, the 10^th term is 5¹⁰ and the n^th term is 5ⁿ.

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

Answer

Given, t₁ = 2 and t_n+1 = 3t_n – 2 for n ≥ 1.

We successively insert the values of n as 1, 2, 3, … and compute the terms.

t₂ = 3t₁ – 2 = 3 × 2 – 2 = 4,

t₃ = 3t₂ – 2 = 3 × 4 – 2 = 10,

t₄ = 3t₃ – 2 = 3 × 10 – 2 = 28,

t₅ = 3t₄ – 2 = 3 × 28 – 2 = 82,

t₆ = 3t₅ – 2 = 3 × 82 – 2 = 244,

t₇ = 3t₆ – 2 = 3 × 244 – 2 = 730.

So, the term 730 occurs at the 7th position.

Hence, 730 is the 7^th term of the sequence.

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

Answer

The given list of numbers 2, 6, 18, … is a GP

first term a = 2

common ratio r = $\dfrac{6}{2}$ = 3.

Let 4374 be the n^th term, or a_n = 4374.

Using the formula, a_n = ar^{n - 1}.

⇒ 4374 = 2(3)^{n - 1}

⇒ $\dfrac{4374}{2} = (3)^{n - 1}$

⇒ 2187 = (3)^{n - 1}

⇒ 3⁷ = (3)^{n - 1}

⇒ n - 1 = 7

⇒ n = 8.

The explicit formula for the n^th term is t_n = ar^{n - 1} = 2 × 3^{n - 1}.

The recursive formula for the n^th term is t₁ = 2, t_n = 3t_{n - 1} for n ≥ 2.

Hence, 4374 is the 8^th term of the GP The explicit formula is t_n = 2 × 3^{n - 1} and the recursive formula is t₁ = 2, t_n = 3t_{n - 1} for n ≥ 2.

A 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. It continues bouncing in this way — each time rising to 60% of the previous height.

(i) What height does the ball reach after the 5^th bounce?

(ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6^th time?

Answer

The ball is dropped from a height of 80 m and rises to 60% (= 0.6) of the previous height after each bounce.

(i) The maximum heights reached after successive bounces are:

1st bounce: 80 × 0.6 = 48 m,

2nd bounce: 48 × 0.6 = 28.8 m,

3rd bounce: 28.8 × 0.6 = 17.28 m,

4th bounce: 17.28 × 0.6 = 10.368 m,

5th bounce: 10.368 × 0.6 = 6.2208 m.

These heights form a GP with first term 48 and common ratio 0.6, so the height after the nth bounce = 80 × (0.6)ⁿ.

∴ Height after the 5th bounce = 80 × (0.6)⁵ = 80 × 0.07776 = 6.2208 m.

Hence, the ball reaches a height of 6.2208 m after the 5^th bounce.

(ii) The ball first falls 80 m and hits the ground for the 1st time. After that, before each bounce, it rises to a height and falls back the same distance. The ball hits the ground for the 6th time after 5 complete up-and-down bounces (the bounces of heights 48, 28.8, 17.28, 10.368 and 6.2208 m).

∴ Total vertical distance = 80 + 2(48 + 28.8 + 17.28 + 10.368 + 6.2208)

= 80 + 2 × 110.6688

= 80 + 221.3376

= 301.3376 m.

Hence, the total vertical distance travelled by the ball by the time it hits the ground for the 6^th time is 301.3376 m.

Which term of the sequence 2, $2\sqrt{2}$, 4, … is 128?

Answer

The given list of numbers is a GP with first term a = 2 and common ratio r = $\dfrac{2\sqrt{2}}{2} = \sqrt{2}.$

Let 128 be the n^th term, or a_n = 128.

By formula, a_n = ar^{n - 1}.

$\Rightarrow 128 = 2(\sqrt{2})^{n - 1} \\[1em] \Rightarrow \dfrac{128}{2} = (2)^{(n - 1)/2} \\[1em] \Rightarrow 64 = (2)^{(n - 1)/2} \\[1em] \Rightarrow 2^6 = (2)^{(n - 1)/2} \\[1em] \Rightarrow \dfrac{n - 1}{2} = 6 \\[1em] \Rightarrow n - 1 = 12 \\[1em] \Rightarrow n = 13.$

Hence, 128 is the 13^th term of the sequence.

Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet. Stage 0 of this fractal is a square sheet of paper. To construct Stage 1, each side of the square is trisected and the points of trisection of opposite sides are joined to obtain nine smaller squares. The centre square is then removed and the 8 smaller squares are retained, leaving a square hole in the centre. The same process is repeated on the eight smaller shaded squares to obtain Stage 2 and so on.

Look at Fig. 8.12 and try to answer the following questions.

(i) How many red squares are there in Stages 0 to 3?

(ii) Can you predict the number of red squares in Stages 4 and 5?

(iii) Can you find a rule for the number of red squares at the n^th stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage.

(iv) Suppose the area of the square in Stage 0 is 1 square unit. What is the area of the red region in Stages 1, 2 and 3? What will be the area of the red region in Stages 4 and 5? Find the explicit as well as the recursive formula for the area of the red region at the n^th stage. What happens to this area as n, the number of stages, goes on increasing?

Answer

At every stage, each red square is divided into 9 smaller squares, the centre one is removed, and 8 red squares are retained. So the number of red squares becomes 8 times the number at the previous stage.

(i) Stage 0: 1 red square,

Stage 1: 1 × 8 = 8 red squares,

Stage 2: 8 × 8 = 64 red squares,

Stage 3: 64 × 8 = 512 red squares.

Hence, the numbers of red squares in Stages 0 to 3 are 1, 8, 64 and 512.

(ii) Continuing the same way (multiplying by 8 each time),

Stage 4: 512 × 8 = 4096 red squares,

Stage 5: 4096 × 8 = 32768 red squares.

Hence, the numbers of red squares at Stages 4 and 5 are 4096 and 32768.

(iii) The number of red squares forms the list 1, 8, 64, 512, …

Here, $\dfrac{8}{1} = \dfrac{64}{8} = \dfrac{512}{64} = 8$, a fixed number, so it is a GP

All these terms are powers of 8:

1 = 8⁰, 8 = 8¹, 64 = 8², 512 = 8³, …

where the exponent of 8 matches the stage number.

The explicit formula for the number of red squares at the n^th stage is t_n = 8ⁿ.

The recursive formula is t₀ = 1, t_n = 8 × t_{n - 1} for n ≥ 1.

Hence, the number of red squares at the n^th stage is t_n = 8ⁿ, with recursive rule t₀ = 1, t_n = 8 × t_{n - 1} for n ≥ 1.

(iv) In each stage, the square is divided into 9 equal smaller squares and 1 (the centre) is removed, so $\dfrac{8}{9}$ of the area remains. The same process is repeated at every stage, so the red area is multiplied by $\dfrac{8}{9}$ each time.

Area at Stage 1 = $1 \times \dfrac{8}{9} = \dfrac{8}{9}$ square units,

Area at Stage 2 = $\dfrac{8}{9} \times \dfrac{8}{9} = \Big(\dfrac{8}{9}\Big)^2 = \dfrac{64}{81}$ square units,

Area at Stage 3 = $\dfrac{64}{81} \times \dfrac{8}{9} = \Big(\dfrac{8}{9}\Big)^3 = \dfrac{512}{729}$ square units.

For Stages 4 and 5,

Area at Stage 4 = $\Big(\dfrac{8}{9}\Big)^4 = \dfrac{4096}{6561}$ square units,

Area at Stage 5 = $\Big(\dfrac{8}{9}\Big)^5 = \dfrac{32768}{59049}$ square units.

The areas form a GP with first term 1 and common ratio $\dfrac{8}{9}$.

The explicit formula for the area of the red region at the n^th stage is s_n = $\Big(\dfrac{8}{9}\Big)^n$.

The recursive formula is s₀ = 1, s_n = $\dfrac{8}{9}$ × s_{n - 1} for n ≥ 1.

As n goes on increasing, the common ratio $\dfrac{8}{9}$ is less than 1, so each term is smaller than the previous one.

Thus, the area of the red region keeps decreasing and gets closer and closer to 0.

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

Answer

Given,

t₁₁ = 38 $\quad$ (Eq 1)

t₁₆ = 73 $\quad$ (Eq 2)

By using formula t_n = a + (n - 1)d for Eq 1 we get,

⇒ t₁₁ = a + (11 - 1)d = 38

⇒ a + 10d = 38 $\quad$ (Eq 3)

By using formula t_n = a + (n - 1)d for Eq 2 we get,

⇒ t₁₆ = a + (16 - 1)d = 73

⇒ a + 15d = 73 $\quad$ (Eq 4)

Subtracting Eq 3 from Eq 4,

⇒ (a + 15d) - (a + 10d) = 73 - 38

⇒ 5d = 35

⇒ d = 7.

Putting value of d in Eq 3,

⇒ a + 10(7) = 38

⇒ a + 70 = 38

⇒ a = 38 - 70

⇒ a = -32.

∴ 31^st term = t₃₁ = a + (31 - 1)d

= -32 + 30(7)

= -32 + 210

= 178.

Hence, the 31^st term of the AP is 178.

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

Answer

Given

t₃ = 16 $\quad$ (Eq 1)

t₇ - t₅ = 12 $\quad$ (Eq 2)

By using formula t_n = a + (n - 1)d on Eq 1 we get,

⇒ t₃ = a + (3 - 1)d = 16

⇒ a + 2d = 16

⇒ a = 16 - 2d. $\quad$ (Eq 3)

By using formula t_n = a + (n - 1)d on Eq 2 we get,

⇒ t₇ - t₅ = 12

⇒ a + (7 - 1)d - [a + (5 - 1)d] = 12

⇒ a + 6d - a - 4d = 12

⇒ 2d = 12

⇒ d = 6.

∴ a = 16 - 2d

= 16 - 2(6)

= 16 - 12

= 4.

t₂ = a + d = 4 + 6 = 10,

t₃ = t₂ + d = 10 + 6 = 16,

t₄ = t₃ + d = 16 + 6 = 22.

Hence, the AP is 4, 10, 16, 22, …

How many three-digit numbers are divisible by 7?
(Hint: All three-digit numbers divisible by 7 form an AP. Find the smallest and largest such three-digit numbers.)

Answer

The three-digit numbers range from 100 to 999.

The smallest three-digit number divisible by 7 is 105 (since 7 × 15 = 105).

The largest three-digit number divisible by 7 is 994 (since 7 × 142 = 994).

So the three-digit numbers divisible by 7 form the AP:

105, 112, 119, …, 994

Here, first term a = 105,

common difference d = 7

last term = l = t_n = 994.

By formula,

t_n = a + (n - 1)d

⇒ 994 = 105 + (n - 1)7

⇒ 994 - 105 = (n - 1)7

⇒ 889 = (n - 1)7

⇒ n - 1 = $\dfrac{889}{7}$

⇒ n - 1 = 127

⇒ n = 128.

Hence, there are 128 three-digit numbers divisible by 7.

How many multiples of 4 lie between 10 and 250?
(Hint: All multiples of 4 form an AP. Find the smallest and largest multiples of 4 between 10 and 250.)

Answer

The smallest multiple of 4 greater than 10 is 12.

The largest multiple of 4 less than 250 is 248.

So the multiples of 4 between 10 and 250 form the AP:

12, 16, 20, …, 248

Here, first term a = 12,

common difference d = 4

last term = l = t_n = 248.

By formula,

t_n = a + (n - 1)d

⇒ 248 = 12 + (n - 1)4

⇒ 248 - 12 = (n - 1)4

⇒ 236 = (n - 1)4

⇒ n - 1 = $\dfrac{236}{4}$

⇒ n - 1 = 59

⇒ n = 60.

Hence, there are 60 multiples of 4 between 10 and 250.

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

Answer

Let the GP be a, ar, ar², ar³, ar⁴, .... with first term a and common ratio r.

Given, sum of the first two terms is -4,

⇒ a + ar = -4

⇒ a(1 + r) = -4 $\quad$ (Eq 1)

Given, the fifth term is 4 times the third term,

⇒ t₅ = 4t₃

⇒ ar⁴ = 4 × ar²

Dividing both sides by ar²,

⇒ r² = 4

⇒ r = ± 2.

Case 1 : When r = 2.

Putting r = 2 in Eq 1,

⇒ a(1 + 2) = -4

⇒ 3a = -4

⇒ a = $-\dfrac{4}{3}$.

The GP is $-\dfrac{4}{3}, -\dfrac{8}{3}, -\dfrac{16}{3}, -\dfrac{32}{3}, ....$

Case 2 : When r = -2.

Putting r = -2 in Eq 1,

⇒ a(1 + (-2)) = -4

⇒ a(-1) = -4

⇒ a = 4.

The GP is 4, -8, 16, -32, ....

Hence, the required GP is $\bold {-\dfrac{4}{3}, -\dfrac{8}{3}, -\dfrac{16}{3}, ....}$ (with r = 2) or 4, -8, 16, -32, .... (with r = -2).

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

Answer

Let 100 be the sum of n consecutive natural numbers starting from a.

These numbers are a, a + 1, a + 2, ....

Let S be the sum of these n consecutive numbers. Writing S in two ways :

$S \phantom{00} = a + (a+1) + (a+2) + \cdots + (a+n-1)$

$S \phantom{00} = (a+n-1) + (a+n-2) + (a+n-3) + \cdots + a$

Adding both rows term by term, each pair adds up to 2a + n - 1, and there are n such pairs :

$2S = n(2a + n - 1)$

$\Rightarrow S = \dfrac{n(2a + n - 1)}{2} = 100$

$\Rightarrow n(2a + n - 1) = 200$

$\Rightarrow 2a + n - 1 = \dfrac{200}{n}$

$\Rightarrow a = \dfrac{1}{2}\Big(\dfrac{200}{n} - n + 1\Big).$

Since a $\geq$ 1, we have 2a + n - 1 > n, so n$^{2}$ < 200, that is, n $\leq$ 14. Also n must divide 200 for a to be a natural number.

The divisors of 200 that are between 2 and 14 are 2, 4, 5, 8 and 10. Checking each :

For n = 2 :

$a = \dfrac{1}{2}\Big(\dfrac{200}{2} - 2 + 1\Big) \\[1em] = \dfrac{1}{2}\Big(\dfrac{200 - 4 + 2}{2}\Big) \\[1em] = \dfrac{1}{2}\Big(\dfrac{198}{2}\Big) \\[1em] = \dfrac{99}{2}, \text{ not a natural number.}$

For n = 4 :

$a = \dfrac{1}{2}\Big(\dfrac{200}{4} - 4 + 1\Big) \\[1em] = \dfrac{1}{2}\Big(\dfrac{200 - 16 + 4}{4}\Big) \\[1em] = \dfrac{1}{2}\Big(\dfrac{188}{4}\Big) \\[1em] = \dfrac{47}{2}, \text{ not a natural number.}$

For n = 5 :

$a = \dfrac{1}{2}\Big(\dfrac{200}{5} - 5 + 1\Big) \\[1em] = \dfrac{1}{2}\Big(\dfrac{200 - 25 + 5}{5}\Big) \\[1em] = \dfrac{1}{2}\Big(\dfrac{180}{5}\Big) \\[1em] = \dfrac{1}{2}(36) \\[1em] = 18, \text{ a natural number.}$

For n = 8 :

$a = \dfrac{1}{2}\Big(\dfrac{200}{8} - 8 + 1\Big) \\[1em] = \dfrac{1}{2}\Big(\dfrac{200 - 64 + 8}{8}\Big) \\[1em] = \dfrac{1}{2}\Big(\dfrac{144}{8}\Big) \\[1em] = \dfrac{1}{2}(18) \\[1em] = 9, \text{ a natural number.}$

For n = 10 :

$a = \dfrac{1}{2}\Big(\dfrac{200}{10} - 10 + 1\Big) \\[1em] = \dfrac{1}{2}\Big(\dfrac{200 - 100 + 10}{10}\Big) \\[1em] = \dfrac{1}{2}\Big(\dfrac{110}{10}\Big) \\[1em] = \dfrac{11}{2}, \text{ not a natural number.}$

So there are two valid ways :

For n = 5, a = 18 : $\quad$ 18 + 19 + 20 + 21 + 22 = 100.

For n = 8, a = 9 : $\quad$ 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100.

Hence, 100 can be expressed as the sum of consecutive natural numbers in two ways : 18 + 19 + 20 + 21 + 22 and 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16.

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 2^nd hour, 4^th hour and n^th hour?

Answer

The number of bacteria doubles every hour, so the number present at successive hours forms a GP.

Originally (at the start), the number of bacteria = 30.

At the end of the 1^st hour = 30 × 2 = 60,
At the end of the 2^nd hour = 60 × 2 = 120,
At the end of the 3^rd hour = 120 × 2 = 240, and so on.

The number present at the end of each hour forms the GP: 60, 120, 240, … with first term 60 and common ratio 2.

In general, the number of bacteria present at the end of the n^th hour is obtained by multiplying the original count by 2 a total of n times:

Number at the end of n^th hour = 30 × 2ⁿ.

At the end of the 2^nd hour = 30 × 2² = 30 × 4 = 120.

At the end of the 4^th hour = 30 × 2⁴ = 30 × 16 = 480.

Hence, the number of bacteria present at the end of the 2^nd hour is 120, at the end of the 4^th hour is 480 and at the end of the n^th hour is 30 × 2ⁿ.

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

Answer

Given,
t₄ + t₈ = 24 $\quad$ (Eq 1)

t₆ + t₁₀ = 44 $\quad$ (Eq 2)

By using formula t_n = a + (n - 1)d for Eq 1 we get,

⇒ t₄ + t₈ = 24

⇒ [a + (4 - 1)d] + [a + (8 - 1)d] = 24

⇒ a + 3d + a + 7d = 24

⇒ 2a + 10d = 24

⇒ a + 5d = 12 $\quad$ (Eq 3)

By using formula t_n = a + (n - 1)d for Eq 2 we get,

⇒ t₆ + t₁₀ = 44

⇒ [a + (6 - 1)d] + [a + (10 - 1)d] = 44

⇒ a + 5d + a + 9d = 44

⇒ 2a + 14d = 44

⇒ a + 7d = 22 $\quad$ (Eq 4)

Subtracting Eq 3 from Eq 4,

⇒ (a + 7d) - (a + 5d) = 22 - 12

⇒ 2d = 10

⇒ d = 5.

Putting value of d in Eq 3,

⇒ a + 5(5) = 12

⇒ a + 25 = 12

⇒ a = 12 - 25

⇒ a = -13.

Hence, the first three terms of the AP are

t₁ = a = -13,

t₂ = a + d = -13 + 5 = -8,

t₃ = a + 2d = -13 + 2(5) = -13 + 10 = -3.

Hence, the first three terms of the AP are -13, -8, -3.

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

Answer

The sum of the first n natural numbers is given by the formula,

S_n = $\dfrac{n(n + 1)}{2}$

We require S_n > 1000,

$\Rightarrow \dfrac{n(n + 1)}{2} \gt 1000 \\[1em] \Rightarrow n(n + 1) \gt 2000.$

Let us check values of n near the required value.

For n = 44:

S₄₄ = $\dfrac{44 \times 45}{2} = \dfrac{1980}{2} = 990,$

which is not greater than 1000.

For n = 45:

S₄₅ = $\dfrac{45 \times 46}{2} = \dfrac{2070}{2} = 1035,$

which is greater than 1000.

Hence, the smallest value of n for which the sum of the first n natural numbers is greater than 1,000 is 45.

Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula for the n^th term.

Answer

The given GP is 2, 8, 32, …

Here, first term a = 2

common ratio r = $\dfrac{8}{2}$ = 4.

Let 131072 be the n^th term of the GP, so t_n = 131072.

By formula,

t_n = ar^{n - 1}

⇒ 131072 = 2(4)^{n - 1}

⇒ 4^{n - 1} = $\dfrac{131072}{2}$

⇒ 4^{n - 1} = 65536

⇒ 4^{n - 1} = 4⁸

⇒ n - 1 = 8

⇒ n = 9.

So 131072 is the 9^th term of the GP.

The explicit formula for the n^th term is

t_n = ar^{n - 1} = 2 × 4^{n - 1}.

The recursive formula for the n^th term is

t₁ = 2, t_n = 4t_{n - 1} for n ≥ 2.

Hence, 131072 is the 9^th term, the explicit formula is t_n = 2 × 4^{n - 1} and the recursive formula is t₁ = 2, t_n = 4t_{n - 1} for n ≥ 2.

The sum of the first three terms of a GP is $\dfrac{13}{12}$ and their product is –1. Find the common ratio and the terms.

Answer

Let the first three terms of the GP be $\dfrac{a}{r}, a, ar.$

Given, their product is -1,

$⇒ \dfrac{a}{r} \times a \times ar = -1 \\[1em] ⇒ a^3 = -1 \\[1em] ⇒ a = -1.$

Given, the sum of the first three terms is $\dfrac{13}{12},$

$\Rightarrow \dfrac{a}{r} + a + ar = \dfrac{13}{12} \\[1em] \Rightarrow a\Big(\dfrac{1}{r} + 1 + r\Big) = \dfrac{13}{12}.$

Putting a = -1,

$\Rightarrow -\Big(\dfrac{1}{r} + 1 + r\Big) = \dfrac{13}{12} \\[1em] \Rightarrow \dfrac{1}{r} + r = -\dfrac{13}{12} - 1 \\[1em] \Rightarrow \dfrac{1 + r^2}{r} = -\dfrac{25}{12} \\[1em] \Rightarrow 12(1 + r^2) = -25r \\[1em] \Rightarrow 12r^2 + 25r + 12 = 0 \\[1em] \Rightarrow 12r^2 + 16r + 9r + 12 = 0 \\[1em] \Rightarrow 4r(3r + 4) + 3(3r + 4) = 0 \\[1em] \Rightarrow (4r + 3)(3r + 4) = 0 \\[1em] \Rightarrow 4r + 3 = 0 \text{ or } 3r + 4 = 0 \\[1em] \Rightarrow r = -\dfrac{3}{4} \text{ or } r = -\dfrac{4}{3}.$

Case 1: When r = $-\dfrac{3}{4}$ and a = -1.

$\dfrac{a}{r} = \dfrac{-1}{-\dfrac{3}{4}} = \dfrac{4}{3}, \\[1em] a = -1, \\[1em] ar = -1 \times \Big(-\dfrac{3}{4}\Big) = \dfrac{3}{4}.$

The terms are $\dfrac{4}{3}, -1, \dfrac{3}{4}.$

Case 2: When r = $-\dfrac{4}{3}$ and a = -1.

$\dfrac{a}{r} = \dfrac{-1}{-\dfrac{4}{3}} = \dfrac{3}{4}, \\[1em] a = -1, \\[1em] ar = -1 \times \Big(-\dfrac{4}{3}\Big) = \dfrac{4}{3}.$

The terms are $\dfrac{3}{4}, -1, \dfrac{4}{3}$ (which is the same set of numbers in reverse order).

Hence, the common ratio is $\bold {-\dfrac{3}{4}}$ or $\bold {-\dfrac{4}{3}}$, and the terms are $\bold {\dfrac{4}{3}, -1, \dfrac{3}{4}}$ or $\bold {\dfrac{3}{4}, -1, \dfrac{4}{3}.}$

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

Answer

Given,

t₄ = x, t₁₀ = y and t₁₆ = z.

We know that

t_n = ar^{n - 1}

∴ t₄ = ar³ = x,

t₁₀ = ar⁹ = y

t₁₆ = ar¹⁵ = z.

Dividing y by x we get,

$\dfrac{y}{x} = \dfrac{ar^9}{ar^3} = r^6$

Dividing z by y we get,

$\dfrac{z}{y} = \dfrac{ar^{15}}{ar^9} = r^6 \\[1em] \therefore \dfrac{y}{x} = \dfrac{z}{y}.$

Since the ratio of the second to the first equals the ratio of the third to the second, x, y, z are in GP.

Hence, proved that x, y, z are in GP.

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.

Answer

Let the first three terms of the GP be a, ar, ar².

Given, the sum of the first three terms is 26,

⇒ a + ar + ar² = 26

⇒ a(1 + r + r²) = 26 $\quad$ (Eq 1)

Given, the sum of their squares is 364,

⇒ a² + a²r² + a²r⁴ = 364

⇒ a²(1 + r² + r⁴) = 364 $\quad$ (Eq 2)

We know that 1 + r² + r⁴ = (1 + r + r²)(1 - r + r²), so Eq 2 becomes

⇒ a²(1 + r + r²)(1 - r + r²) = 364 $\quad$ (Eq 3)

Dividing Eq 3 by Eq 1,

$\Rightarrow \dfrac{a^2(1 + r + r^2)(1 - r + r^2)}{a(1 + r + r^2)} = \dfrac{364}{26} \\[1em] \Rightarrow a(1 - r + r^2) = 14. \quad \text{ (Eq 4)}$

Dividing Eq 1 by Eq 4,

$\Rightarrow \dfrac{a(1 + r + r^2)}{a(1 - r + r^2)} = \dfrac{26}{14} \\[1em] \Rightarrow \dfrac{1 + r + r^2}{1 - r + r^2} = \dfrac{13}{7} \\[1em] \Rightarrow 7(1 + r + r^2) = 13(1 - r + r^2) \\[1em] \Rightarrow 7 + 7r + 7r^2 = 13 - 13r + 13r^2 \\[1em] \Rightarrow 6r^2 - 20r + 6 = 0 \\[1em] \Rightarrow 3r^2 - 10r + 3 = 0 \\[1em] \Rightarrow 3r^2 - 9r - r + 3 = 0 \\[1em] \Rightarrow 3r(r - 3) - 1(r - 3) = 0 \\[1em] \Rightarrow (3r - 1)(r - 3) = 0 \\[1em] \Rightarrow r = \dfrac{1}{3} \text{ or } r = 3.$

Case 1 : When r = 3.

Putting r = 3 in Eq 1,

⇒ a(1 + 3 + 3²) = 26

⇒ a(1 + 3 + 9) = 26

⇒ 13a = 26

⇒ a = 2.

The terms are 2, 2 × 3, 2 × 3², that is, 2, 6, 18.

Case 2 : When r = $\dfrac{1}{3}$.

Putting r = $\dfrac{1}{3}$ in Eq 1,

$\Rightarrow a\Big(1 + \dfrac{1}{3} + \dfrac{1}{9}\Big) = 26 \\[1em] \Rightarrow a\Big(\dfrac{9 + 3 + 1}{9}\Big) = 26 \\[1em] \Rightarrow \dfrac{13a}{9} = 26 \\[1em] \Rightarrow a = 18.$

The terms are 18, 6, 2 (which is the same set of numbers in reverse order).

Hence, the required GP is 2, 6, 18 with common ratio 3, or 18, 6, 2 with common ratio $\dfrac{1}{3}$.

Suppose P₁ = 1, P₂ = 2 and for n > 2, P_n = P₁ + P₂ + … + P_n–1 + 1. Find the values of P₁, P₂, …, P₈. Can you find a simpler recursive formula for P_n? Can you give an explicit formula?

Answer

Given, P₁ = 1, P₂ = 2 and for n > 2, P_n = P₁ + P₂ + … + P_n–1 + 1.

Computing successively,

P₁ = 1,

P₂ = 2,

P₃ = P₁ + P₂ + 1 = 1 + 2 + 1 = 4,

P₄ = P₁ + P₂ + P₃ + 1 = 1 + 2 + 4 + 1 = 8,

P₅ = P₁ + P₂ + P₃ + P₄ + 1 = 1 + 2 + 4 + 8 + 1 = 16,

P₆ = 1 + 2 + 4 + 8 + 16 + 1 = 32,

P₇ = 1 + 2 + 4 + 8 + 16 + 32 + 1 = 64,

P₈ = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 1 = 128.

So the values are P₁, P₂, …, P₈ = 1, 2, 4, 8, 16, 32, 64, 128.

Simpler recursive formula:

For n > 2,
P_n = (P₁ + P₂ + … + P_n–2 + P_n–1) + 1, and

P_n–1 = (P₁ + P₂ + … + P_n–2) + 1.

Subtracting the second from the first,

⇒ P_n - P_n–1 = P_n–1

⇒ P_n = 2P_n–1.

This also holds for P₂ = 2P₁ = 2. So the simpler recursive formula is

P₁ = 1, P_n = 2P_n–1 for n ≥ 2.

Explicit formula:

Since each term is twice the previous one with P₁ = 1, the sequence 1, 2, 4, 8, … consists of powers of 2:

P_n = 2^{n - 1}.

Hence, P₁, …, P₈ = 1, 2, 4, 8, 16, 32, 64, 128, the simpler recursive formula is P₁ = 1, P_n = 2P_n–1 for n ≥ 2, and the explicit formula is P_n = 2^{n - 1}.

Suppose W₁ = 1, W₂ = 2 and for n > 2, W_n = W₁ + W₂ + … + W_n–2 + 2. Find the values of W₁, W₂, …, W₈. Do you recognise this sequence?

Answer

Given, W₁ = 1, W₂ = 2 and for n > 2, W_n = W₁ + W₂ + … + W_n–2 + 2.

Computing successively,

W₁ = 1,

W₂ = 2,

W₃ = W₁ + 2 = 1 + 2 = 3,

W₄ = W₁ + W₂ + 2 = 1 + 2 + 2 = 5,

W₅ = W₁ + W₂ + W₃ + 2 = 1 + 2 + 3 + 2 = 8,

W₆ = W₁ + W₂ + W₃ + W₄ + 2 = 1 + 2 + 3 + 5 + 2 = 13,

W₇ = W₁ + W₂ + W₃ + W₄ + W₅ + 2 = 1 + 2 + 3 + 5 + 8 + 2 = 21,

W₈ = W₁ + W₂ + W₃ + W₄ + W₅ + W₆ + 2 = 1 + 2 + 3 + 5 + 8 + 13 + 2 = 34.

So the values are W₁, W₂, …, W₈ = 1, 2, 3, 5, 8, 13, 21, 34.

We can also find a simpler relation. For n > 2,

W_n = (W₁ + W₂ + … + W_n–3 + W_n–2) + 2, and

W_n–1 = (W₁ + W₂ + … + W_n–3) + 2.

Subtracting the second from the first,

⇒ W_n - W_n–1 = W_n–2

⇒ W_n = W_n–1 + W_n–2.

So each term is the sum of the previous two terms. With W₁ = 1 and W₂ = 2, the sequence 1, 2, 3, 5, 8, 13, 21, 34, … is exactly the Virahānka–Fibonacci sequence.

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