Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.2 Solutions Predicting What Comes Next: Exploring Sequences and Progressions

Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.2 Solutions for Predicting What Comes Next: Exploring Sequences and Progressions provides step-by-step solutions to the questions given in the exercise. Students can use these solutions to understand patterns, identify sequences and explore how terms in a progression are formed.

The detailed explanations make it easier to understand the concepts and apply the appropriate methods while solving problems. These solutions are useful for homework, revision, exam preparation and strengthening mathematical problem-solving skills.

Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.2 Solutions Predicting What Comes Next: Exploring Sequences and Progressions

Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.2 Solutions Predicting What Comes Next: Exploring Sequences and Progressions

NCERT Solutions for Class 9 Maths Chapter 8 Exercise Set 8.2

Exercise Set 8.2 focuses on arithmetic progressions, nth term of an AP and the sum of first n natural numbers.

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

Answer:
First term:

a = 3

Common difference:

d = 8 - 3 = 5

nth term of an AP:

tₙ = a + (n - 1)d

For n = 10:

t₁₀ = 3 + (10 - 1)5
= 3 + 45
= 48

For n = 26:

t₂₆ = 3 + (26 - 1)5
= 3 + 125
= 128

Final answer:
10th term = 48, 26th term = 128

Question 2. Which term of the AP 21, 18, 15, … is -81? Also, is 0 a term of this AP?

Answer:
First term:

a = 21

Common difference:

d = 18 - 21 = -3

nth term:

tₙ = 21 + (n - 1)(-3)
= 21 - 3n + 3
= 24 - 3n

For -81:

24 - 3n = -81
-3n = -105
n = 35

So, -81 is the 35th term.

Now check 0:

24 - 3n = 0
3n = 24
n = 8

Since n is a natural number, 0 is also a term.

Final answer:
-81 is the 35th term, and 0 is the 8th term.

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

Answer:
First term:

a = 11

Common difference:

d = 8 - 11 = -3

nth term:

tₙ = a + (n - 1)d
= 11 + (n - 1)(-3)
= 11 - 3n + 3
= 14 - 3n

Recursive rule:

t₁ = 11
tₙ = tₙ₋₁ - 3, for n ≥ 2

Final answer:
tₙ = 14 - 3n; recursive rule: t₁ = 11, tₙ = tₙ₋₁ - 3

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

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

3rd term:

a + 2d = 12 …(1)

50th term:

a + 49d = 106 …(2)

Subtract (1) from (2):

47d = 94
d = 2

Substitute in (1):

a + 2(2) = 12
a + 4 = 12
a = 8

29th term:

t₂₉ = a + 28d
= 8 + 28(2)
= 8 + 56
= 64

Final answer:
29th term = 64

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

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

12, 15, 18, …, 99

This is an AP with:

a = 12
d = 3
last term = 99

Find n:

99 = 12 + (n - 1)3
87 = 3(n - 1)
n - 1 = 29
n = 30

So, there are 30 such numbers.

Sum:

S = n/2 × (first term + last term)

= 30/2 × (12 + 99)
= 15 × 111
= 1665

Final answer:
There are 30 two-digit numbers divisible by 3, and their sum is 1665.

Question 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?

Answer:
Initial salary = ₹5,00,000
Annual increment = ₹20,000

Let the salary reach ₹7,00,000 after n years.

Increase needed:

₹7,00,000 - ₹5,00,000 = ₹2,00,000

Number of increments:

2,00,000 / 20,000 = 10

Final answer:
Harish’s income reached ₹7,00,000 after 10 years.

Question 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?

Answer:
Total marbles:

1 + 2 + 3 + … + 25

Sum of first n natural numbers:

Sₙ = n(n + 1)/2

For n = 25:

S₂₅ = 25(26)/2
= 25 × 13
= 325

Final answer:
325 marbles

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