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

Given AP: 3, 8, 13, 18, ...

First term, a = 3

Common difference, d = 8 − 3 = 5

Formula: t_n = a + (n − 1)d

10^th Term

t₁₀ = 3 + (10 − 1)(5)

= 3 + 45

= 48

Answer: t₁₀ = 48

26^th Term

t₂₆ = 3 + (26 − 1)(5)

= 3 + 125

= 128

Answer: t₂₆ = 128

Given AP: 21, 18, 15, ...

a = 21

d = −3

t_n = a + (n − 1)d

t_n = 21 + (n − 1)(−3)

t_n = 24 − 3n

For −81

24 − 3n = −81

−3n = −105

n = 35

Answer: −81 is the 35^th term.

Checking Whether 0 is a Term

24 − 3n = 0

3n = 24

n = 8

Since n is a whole number, 0 is a term of the AP.

Answer: 0 is the 8^th term.

Given AP: 11, 8, 5, 2, ...

a = 11

d = −3

n^th Term

t_n = a + (n − 1)d

= 11 + (n − 1)(−3)

= 11 − 3n + 3

= 14 − 3n

Answer:

t_n = 14 − 3n

Recursive Rule

t₁ = 11

t_n+1 = t_n − 3

Recursive Rule:

t₁ = 11, t_n+1 = t_n − 3

Number of terms = 50

3^rd term = 12

Last term = 106

a + 2d = 12

a + 49d = 106

Subtracting:

47d = 94

d = 2

a + 2(2) = 12

a = 8

29^th Term

t₂₉ = a + 28d

= 8 + 28(2)

= 8 + 56

= 64

Answer: t₂₉ = 64

Smallest 2-digit multiple of 3 = 12

Largest 2-digit multiple of 3 = 99

AP: 12, 15, 18, ..., 99

a = 12, d = 3, l = 99

Number of Terms

99 = 12 + (n − 1)3

87 = 3(n − 1)

29 = n − 1

n = 30

Answer: 30 numbers

Sum of All Numbers

S_n = n/2 [a + l]

= 30/2 (12 + 99)

= 15 × 111

= 1665

Answer: Sum = 1665

Initial salary = ₹5,00,000

Annual increment = ₹20,000

Target salary = ₹7,00,000

AP: 5,00,000, 5,20,000, 5,40,000, ...

a = 5,00,000

d = 20,000

t_n = 7,00,000

7,00,000 = 5,00,000 + (n − 1)(20,000)

2,00,000 = (n − 1)(20,000)

10 = n − 1

n = 11

The 11^th salary is ₹7,00,000.

Years required = 10

Answer: After 10 years.

Rows contain: 1, 2, 3, 4, ..., 25

This is an AP with:

a = 1

d = 1

n = 25

Total Marbles

S₂₅ = 25/2 [2(১৷) + (25 − 1)(1)]

= 25/2 [2 + 24]

= 25/2 × 26

= 25 × 13

= 325

Answer: The child uses 325 marbles in all.