Predicting What Comes Next: Exploring Sequences and Progressions Class 9 Maths Ganita Manjari Part 1 Chapter 8 NCERT Solutions

Predicting What Comes Next: Exploring Sequences and Progressions Class 9 Maths Ganita Manjari Part 1 Chapter 8 NCERT Solutions Looking for the best Predicting What Comes Next: Exploring Sequences and Progressions Class 9 Maths Ganita Manjari Part 1 Chapter 8 NCERT Solutions? At SchoolLearners, we provide accurate, easy-to-understand, and step-by-step NCERT solutions that help students prepare for school exams and strengthen their concepts.

Exercise Set 8.1

1. Find the first five terms of the sequence in which the nth term is given by (i) t_n = 3n – 4, (ii) t_n = 2 – 5n, and (iii) t_n = n² – 2n + 3 for n ≥ 1.
Solution:
(i) t_n = 3n – 4
For n = 1, 2, 3, 4, 5:
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
Therefore, the first five terms of the sequence are –1, 2, 5, 8, 11.

(ii) t_n = 2 – 5n
For n = 1, 2, 3, 4, 5:
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
Therefore, the first five terms of the sequence are –3, –8, –13, –18, –23.

(iii) t_n = n² – 2n + 3
For n = 1, 2, 3, 4, 5:
t₁ = (1)² – 2(1) + 3 = 1 – 2 + 3 = 4 – 2 = 2
t₂ = (2)² – 2(2) + 3 = 4 – 4 + 3 = 7 – 4 = 3
t₃ = (3)² – 2(3) + 3 = 9 – 6 + 3 = 12 – 6 = 6
t₄ = (4)² – 2(4) + 3 = 16 – 8 + 3 = 19 – 8 = 11
t₅ = (5)² – 2(5) + 3 = 25 – 10 + 3 = 28 – 10 = 18
Therefore, the first five terms of the sequence are 2, 3, 6, 11, 18.

2. Find the 10th and 15th terms of the sequence t_n = 5n – 3 for n ≥ 1.
Solution:
10th term (n = 10):
t₁₀ = 5(10) – 3 = 50 – 3 = 47
15th term (n = 15):
t₁₅ = 5(15) – 3 = 75 – 3 = 72

3. Determine whether 97 and 172 are terms of the sequence t_n = 5n – 3 for n ≥ 1.
Solution:
(i) tₙ = 5n − 3
For 97 to be a term of the sequence, tₙ = 97
5n − 3 = 97
5n = 97 + 3
5n = 100
n = $\frac{100}{5}$ = 20
Since 20 is a natural number, we conclude that 97 is the 20th term of the sequence.

(ii) For 172 to be a term of the sequence, tₙ = 172
5n − 3 = 172
5n = 172 + 3
5n = 175
n = $\frac{175}{5}$ = 35
Since 35 is a natural number, we conclude that 172 is the 35th term of the sequence.

4. Which term of the sequence t_n = 5n – 3 for n ≥ 1 is 607?
Solution:
For 607 to be a term of the sequence, tₙ = 607
5n – 3 = 607
5n = 607 + 3
5n = 610
n = $\frac{610}{5}$ = 122
Since 122 is a natural number, we conclude that 607 is the 122nd term of the sequence.

5. 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?
Solution:
t₁ = –5
t_{n + 1} = t_n + 3
Substituting n = 1, 2, 3, 4:
t₂ = t₁ + 3 = –5 + 3 = –2
t₃ = t₂ + 3 = –2 + 3 = 1
t₄ = t₃ + 3 = 1 + 3 = 4
t₅ = t₄ + 3 = 4 + 3 = 7
Hence, the first five terms are −5, −2, 1, 4, 7.
This is an arithmetic sequence with first term −5 and common difference 3.
tₙ = −5 + (n − 1) × 3 = −5 + 3n − 3 = 3n − 8
For 52 to be a term of the sequence, tₙ = 52.
52 = 3n – 8
3n = 52 + 8
3n = 60
n = 20
Since 20 is a natural number, wew can conclude that 52 is the 20th term of the sequence.

6. 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 .
Solution:
Given that T₁ = 1, T₂ = 2, T₃ = 4
T_n = T_{n – 1} + T_{n – 2} + T_{n –3}
Now,
T₄​ = T₃ ​+ T₂​ + T_1​= 4 + 2 + 1 = 7
T₅​ = T_4​ + T₃​ + T₂​ = 7 + 4 + 2 = 13
T₆​ = T₅​ + T₄​ + T₃​ = 13 + 7 + 4 = 24
T₇​ = T₆ ​+ T₅ ​+ T_4​ = 24 + 13 + 7 = 44
T₈​ = T_7​ + T₆​ + T₅​ = 44 + 24 + 13 = 81
Therefore, T₄​ = 7, T₅ ​= 13, T₆​ = 24, T₇ ​= 44, T₈ ​= 81.

Exercise Set 8.2

1. Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ….
Solution:
Given AP: 3, 8, 13, 18, …
Here, a = 3, d = 8 − 3 = 5
The nth term of an AP is given by:
a_n​ = a + (n − 1)d
(i) a₁₀​ = 3 + (10 − 1)5
= 3 + 9(5) = 3 + 45 = 48.
(ii) a₂₆​ = 3 + (26 − 1)5
= 3 + 25(5) = 3 + 125 = 128.
Hence, the 10th term is 48, and the 26th term is 128.

2. Which term of the AP: 21, 18, 15,…is –81? Also, is 0 a term of this AP? Give reasons for your answer.
Solution:
Given AP: 21, 18, 15, …
Here, a = 21, d = 18 − 21 = −3
The nth term of an AP is given by:
a_n​ = a + (n − 1)d
(i) Let a_n = –81
– 81 = 21 + (n – 1)(– 3)
– 81 = 21 – 3n + 3
3n = 21 + 81 + 3
3n = 105
n = $\frac{105}{3}$ = 35.
Hence, –81 is the 35th term of the AP.
(ii) Let a_n = 0
0 = 21 + (n – 1)(– 3)
0 = 21 – 3n + 3
3n = 24
n = $\frac{24}{3}$ = 8.
Since 8 is a whole number, 0 is a term of the AP.

3. Find the nth term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.
Solution:
Given AP: 11, 8, 5, 2,…
Here, a = 11, d = 8 − 11 = −3
The nth term of an AP is given by:
a_n​ = a + (n − 1)d
So,
a_n​ = 11 + (n − 1)(– 3)
= 11 – 3n + 3
= 14 – 3n
Hence, the nth term of the AP is (14 – 3n).
Therefore, the recursive rule for this AP is:
t₁ = 11, t_n = t_{n – 1} – 3, n ≥ 2.

4. An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th 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’.)
Solution:
Let the first term be a, and the common difference be d.
a₃ = 12 ………. (Given)
∴ a + 2d = 12 ……. (1)
a₅₀ = 106 ……….. (Given)
∴ a + 49d = 106 ………. (2)
Subtracting (1) from (2),
a + 49d – (a + 2d) = 106 – 12
a + 49d – a – 2d = 94
47d = 94
d = $\frac{94}{47}$ = 2
Substituting d = 2 in (1),
a + 2(2) = 12
a + 4 = 12
a = 12 – 4 = 8
Now,
a₂₉ = 8 + (29 – 1)(2)
= 8 + 28(2)
= 8 + 56 = 64.
Hence, the 29th term is 64.

5. How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?
Solution:
The 2-digit numbers divisible by 3 are:
12, 15, 18, 21, …, 99.
This forms an AP with a = 12, d = 3
Let the number of terms be n.
Using:
a_n​ = a + (n − 1)d
99 = 12 + (n – 1)3
99 = 12 + 3n – 3
3n = 99 – 12 + 3
3n = 90
n = $\frac{90}{3}$ = 30.
Hence, there are 30 two-digit numbers divisible by 3.
Let S be the sum of all these 2-digit numbers.
Thus S = 12 + 15 + 18 + 21 + …..…+ 90 + 93 + 96 + 99.
Also, S = 99 + 96 + 93 + 90 + ……….+ 21 + 18 + 15 + 12.
Adding the two equations for S, we get
2S = 111, 111, 111, 111 + …….. + 111 + 111 + 111+ 111(111 added 30 times)
2S = 3330
S = $\frac{3330}{2}$ = 1665
Therefore, the sum of all the 2-digit numbers divisible by 3 is 1665.

6. 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?
Solution:
Initial salary = ₹5,00,000
Salary after 1st increment = ₹5,00,000 + ₹20,000 = ₹5,20,000
Salary after 2nd increment = ₹5,20,000 + ₹ 20,000 = ₹5,40,000
Salary after 3rd increment = ₹5,40,000 + ₹20,000 = ₹5,60,000
Thus, Harish’s annual salary forms an AP:
500000, 520000, 540000, …
Here,
a = 500000, d = 20000
Let the income become ₹7,00,000 in the nth year.
Using
a_n​ = a + (n − 1)d
700000 = 500000 + (n − 1) × 20000
700000 = 500000 + 20000n − 20000
700000 = 480000 + 20000n
220000 = 20000n
n = $\frac{220000}{20000}$ = 11.
Hence, Harish’s income reached ₹7,00,000 in the 11th year.

7. 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?
Solution:
Number of marbles in the 1st row = 1
Number of marbles in the 2nd row = 2
Number of marbles in the 3rd row = 3
……..
Number of marbles in the 25th row = 25
Thus, the marbles arranged in rows form an AP:
1, 2, 3, …, 25
Since the sum of the first n natural numbers = $\frac{n(n+1)}{2}$
Therefore,
1, 2, 3, …, 25 = $\frac{25\times (25+1)}{2}$ = $\frac{25\times 26}{2}$ = 25 × 13 = 325.
Hence, the child uses 325 marbles in all.

Exercise Set 8.3

1. Find the 12th term of a GP with common ratio 2, whose 8th term is 192.
Solution:
Let the first term be a and the common ratio be r.
r = 2 and t₈ = 192 ……… (Given)
The nth term of the GP is t_n = ar^n–1
So, 192 = a(2)^8–1
192 = a(2)⁷
192 = 128a ………. (∵ 2⁷ = 128)
a = $\frac{192}{128}$ = $\frac{3}{2}$.
Thus,
t₁₂ = $\frac{3}{2}$ × (2)^12–1
= $\frac{3}{2}$ × (2)¹¹
= $\frac{3}{2}$ × 2048 = 3072.
Hence, the 12th term of the GP is 3072.

2. Find the 10th and nth terms of the GP: 5, 25, 125, … .
Solution:
Given GP: 5, 25, 125, …
Here,
a = 5, r = $\frac{25}{5}$= 5
The nth term of the GP is t_n = ar^n–1
(i) t₁₀ = 5 × (5)^10–1
= 5 × (5)⁹ = (5)¹⁰ = 9765625
Hence, the 10th term is 9765625.
(ii) t_n = 5 × (5)^n–1 = 5ⁿ
Hence, the nth term of the GP is 5ⁿ.

3. 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?
Solution:
Given, t₁ ​= 2, t_n+1​ = 3t_n​ − 2, n ≥ 1
Writing the first few terms,
t₁ ​= 2
t₂ ​= 3(2) – 2 = 6 – 2 = 4
t₃ ​= 3(4) – 2 = 12 – 2 = 10
t₄ ​= 3(10) – 2 = 30 – 2 = 28
From the pattern, the nth term is t_n = 3^{n – 1} + 1
Let t_n = 730
So, 730 = 3^{n – 1} + 1
730 – 1 = 3^{n – 1}
729 = 3^{n – 1}
3⁶ = 3^{n – 1}
Comparing exponents,
n – 1 = 6
n = 6 + 1 = 7
Therefore, the 7th term of the sequence is 730.

4. Which term of the GP: 2, 6, 18, … is 4374? Write the explicit formula as well as the recursive formula for the nth term.
Solution:
Given GP: 2, 6, 18, …
Here, a = 2, r = $\frac{6}{2}$ = 3
The nth term of the GP is given by t_n = ar^n–1
Let t_n = 4374
4374 = 2(3)^n–1
3^n–1 = $\frac{4374}{2}$
3^n–1 = 2187
3^n–1 = 3⁷
Comparing exponents,
n – 1 = 7
n = 8
Hence, 4374 is the 8th term of the GP.
The explicit formula for the GP is:
t_n​ = 2(3)ⁿ⁻¹
The recursive formula for the GP is:
t₁​ = 2, t_n+1​ = 3t_n​ , n ≥ 1

5. 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 5th bounce?
(ii)  What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?
Solution:
After each bounce, the ball rises to 60% of its previous height.
Initial height of the ball = 80 m
After 1st bounce = 60% of 80 m
= $\frac{60}{100}$ × 80 = 48 m
After 2nd bounce = 60% of 48 m
= $\frac{60}{100}$ × 48 = 28.8 m
After 3rd bounce = 60% of 28.8 m
= $\frac{60}{100}$ × 28.8 = = $\frac{60}{100}$ × $\frac{288}{10}$ = 17.28 m
Thus, the heights reached after each bounce form a GP:
80, 48, 28.8, 17.28, …
Here, a = 48, r = $\frac{28.8}{48}$ = 0.6
(i) The nth term of the GP is given by t_n = ar^n–1
So,
t₅ = 8(0.6)^5–1
= 8(0.6)⁴
= 48(0.1296) = 6.2208
Hence, the ball reaches a height of 6.2208 m after the 5th bounce.
(ii) The total distance is the sum of the first 6 terms of a geometric series:
80 + 2 × (48 + 28.8 + 17.28 + 10.368 + 6.2208) = 301.3376 m.
Hence, the total vertical distance travelled by the time the ball hits the ground for the 6th time is 301.34 m.

6. Which term of the sequence 2, 2$\sqrt{2}$, 4, … is 128?
Solution:
Here, a = 2, r = $\frac{2\sqrt{2}}{2}$ = $\sqrt{2}$
The nth term of the GP is given by t_n = ar^n–1
Since t_n = 128
128 = 2$(\sqrt{2})$^n–1
$\frac{128}{2}$ = $(\sqrt{2})$^{n – 1}
64 = $(\sqrt{2})$^{n – 1}
2⁶ = ${\left(2^{1/2}\right)}^{n-1}$
2⁶ = $2^{\frac{n-1}{2}}$
Comparing powers,
$\frac{n-1}{2}$ = 6
n – 1 = 12
n = 12 + 1 = 13.
Therefore, 128 is the 13th term of the sequence.

7. 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 nth 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 nth stage. What happens to this area as n, the number of stages, goes on increasing?
Solution:
(i) Stage 0 = 1 square
Stage 1 = 8 squares
Stage 2 = 8 × 8 = 64 squares
Stage 3 = 64 × 8 = 512 squares
Hence, the number of red squares in Stages 0 to 3 = 1 + 8 + 64 + 512 = 585 squares.

(ii) Each stage has 8 times the number of red squares of the previous stage.
Stage 4 = 512 × 8 = 4096 squares
Stage 5 = 4096 × 8 = 32768 squares
Therefore, the number of red squares at Stage 4 is 4096, and at Stage 5 is 32768.

(iii) The sequence is: 1, 8, 64, 512, …
This is a G.P. with a = 1 and r = 8
Explicit formula: t_n​ = arⁿ
So, t_n​ = 8ⁿ
Recursive formula:
T₀​ = 1
Tn​ = 8Tⁿ⁻¹​, n ≥ 1

(iv) Area of Stage 0 = 1 sq. unit
At each stage, ${\displaystyle \frac{1}{9}}$​ part is removed, so ${\displaystyle \frac{8}{9}}$​ remains.
Area in Stage 1 = $\frac{8}{9}$ sq. unit
Area in Stage 2 = ${\left(\frac{8}{9}\right)}^2$sq. unit
Area in Stage 3 = ${\left(\frac{8}{9}\right)}^3$sq. unit
Area in Stage 4 = ${\left(\frac{8}{9}\right)}^4$sq. unit
Area in Stage 5 = ${\left(\frac{8}{9}\right)}^5$sq. unit
Thus, areas form a G.P.: 1, $\frac{8}{9}$, ${\left(\frac{8}{9}\right)}^2$, ${\left(\frac{8}{9}\right)}^3$, ${\left(\frac{8}{9}\right)}^4$, ${\left(\frac{8}{9}\right)}^5$, ……
Recursive formula:
A₀ = 1, A_n = $\frac{8}{9}$A_{n – 1}
As n increases, the area keeps decreasing and approaches 0.

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