Class 9 - Chapter 8: Predicting What Comes Next: Exploring Sequences and Progressions

Given:

T₁₁ = 38

T₁₆ = 73

Using T_n = a + (n − 1)d

a + 10d = 38 ..........(1)

a + 15d = 73 ..........(2)

Subtracting (1) from (2):

5d = 35

d = 7

Substituting in (1):

a + 70 = 38

a = −32

T₃₁ = a + 30d

= −32 + 30(7)

= −32 + 210

= 178

Answer: T₃₁ = 178

Given:

3^rd term = 16

7^th term exceeds 5^th term by 12

a + 2d = 16 ..........(1)

(a + 6d) − (a + 4d) = 12

2d = 12

d = 6

Substituting into (1):

a + 12 = 16

a = 4

AP:

4, 10, 16, 22, 28, ...

Smallest 3-digit multiple of 7 = 105

Largest 3-digit multiple of 7 = 994

AP: 105, 112, 119, ..., 994

a = 105, d = 7

994 = 105 + (n − 1)7

889 = 7(n − 1)

127 = n − 1

n = 128

Answer: 128 numbers

Smallest multiple = 12

Largest multiple = 248

AP: 12, 16, 20, ..., 248

248 = 12 + (n − 1)4

236 = 4(n − 1)

59 = n − 1

n = 60

Answer: 60 multiples

First term = a

Common ratio = r

First two terms sum = −4

a + ar = −4

a(1+r) = −4 ..........(1)

5^th term = 4 × 3^rd term

ar⁴ = 4ar²

r² = 4

r = ±2

Taking r = 2:

a(3) = −4

a = −4/3

One GP:

−4/3, −8/3, −16/3, ...

100 = 18 + 19 + 20 + 21 + 22

100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16

Possible expressions:

• 18 + 19 + 20 + 21 + 22 = 100
• 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100

Initial bacteria = 30

Doubles every hour

GP: 30, 60, 120, 240, ...

a = 30

r = 2

After 2 Hours

30 × 2² = 120

After 4 Hours

30 × 2⁴ = 480

After n Hours

T = 30 × 2ⁿ

Answers:

• 2 Hours = 120
• 4 Hours = 480
• n Hours = 30 × 2ⁿ

T₄ + T₈ = 24

(a+3d)+(a+7d)=24

2a+10d=24

a+5d=12 ..........(1)

T₆ + T₁₀ = 44

(a+5d)+(a+9d)=44

2a+14d=44

a+7d=22 ..........(2)

Subtracting:

2d=10

d=5

a+25=12

a=−13

First three terms:

−13, −8, −3

Answer: −13, −8, −3

S_n = n(n+1)/2

n(n+1)/2 > 1000

n(n+1) > 2000

Checking values:

44 × 45 = 1980

45 × 46 = 2070

2070 > 2000

Answer: n = 45

GP: 2, 8, 32, ...

a = 2

r = 4

T_n = 2(4ⁿ⁻¹)

131072 = 2(4ⁿ⁻¹)

65536 = 4ⁿ⁻¹

65536 = 2¹⁶

4⁸ = 65536

n − 1 = 8

n = 9

Answer: 131072 is the 9^th term.

Explicit Formula

T_n = 2(4ⁿ⁻¹)

Recursive Formula

T₁ = 2

T_n+1 = 4T_n

Terms: a/r, a, ar

Product = −1

(a/r)(a)(ar)=a³=−1

a = −1

Sum = 13/12

−(1/r + 1 + r)=13/12

1/r + 1 + r = −13/12

12r + 12r² + 12 = −13r

12r² + 25r + 12 = 0

(3r+4)(4r+3)=0

r = −4/3 or r = −3/4

Terms:

3/4, −1, 4/3

or

4/3, −1, 3/4

Let GP be:

a, ar, ar², ar³, ...

Then:

x = ar³

y = ar⁹

z = ar¹⁵

y² = (ar⁹)²

= a²r¹⁸

xz = (ar³)(ar¹⁵)

= a²r¹⁸

Therefore:

y² = xz

Hence x, y and z are in GP.

Let terms be:

a/r, a, ar

3a = 26

a = 26/3

After solving,

r = 3/2

Terms:

52/9, 26/3, 13

P₁=1

P₂=2

n	Pn
1	1
2	2
3	4
4	8
5	16
6	32
7	64
8	128

Pattern: P_n = 2ⁿ⁻¹

Recursive: P_n = 2P_n−1

W₁=1

W₂=2

n	Wn
1	1
2	2
3	5
4	10
5	20
6	40
7	80
8	160

Answer:

W₁=1, W₂=2, W₃=5, W₄=10, W₅=20, W₆=40, W₇=80, W₈=160

From W₄ onwards, each term doubles.