Question 1
↑ Jump to NavigatorFind the 10th and 26th terms of the AP: 3, 8, 13, 18, ….
Solution
Given:
We know,
For the 10th term:
t₁₀ = 3 + (10 − 1) × 5
For the 26th term:
t₂₆ = 3 + (26 − 1) × 5
Therefore:
Answer:t₁₀ = 48 and t₂₆ = 128
Chapter 8: Predicting What Comes Next: Exploring Sequences and ProgressionsPage 185–186All 7 Questions on This Page
Class 9 Maths Chapter 8 Exercise Set 8.2 Solutions
Complete step-by-step solutions for Class 9 Mathematics Ganita Manjari Chapter 8 Exercise Set 8.2. Covers finding nth terms of an AP, identifying term positions, recursive rules, solving simultaneous linear equations for AP parameters, and AP sum applications.
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Key Concepts for Exercise Set 8.2
Jump to Question 11
Arithmetic Progression (AP)
A sequence of numbers where the difference between any consecutive terms remains constant throughout.
2
First Term (a)
The very first term of the arithmetic progression, denoted by a or t₁.
t₁ = a
3
Common Difference (d)
The constant difference added to each term to obtain the next term (d = tₙ − tₙ₋₁). Can be positive, negative, or zero.
d = tₙ − tₙ₋₁
4
General nth Term (tₙ)
Used to determine any term position in an AP without writing out all preceding terms:
tₙ = a + (n − 1)d
5
Finding Term Position (n)
To find which term has a given value, set tₙ equal to that value and solve the linear equation for n. The value of n must always be a positive integer (n ∈ {1, 2, 3, …}).
6
Recursive Rule for an AP
Expresses each term based on the value of the term immediately before it:
t₁ = a, tₙ = tₙ₋₁ + d, for n ≥ 2
7
Finding a and d from Given Terms
When two distinct terms are known (e.g. t₃ and t₅₀), form a pair of simultaneous linear equations using tₙ = a + (n − 1)d and solve for a and d.
t₃: a + 2d = 12, t₅₀: a + 49d = 106
8
Sum of First n Terms (Sₙ)
When the first term a and the last term l are known, the sum of all n terms is given by:
Sₙ = n/2 (a + l)
Question 2
↑ Jump to NavigatorWhich term of the AP : 21, 18, 15, … is – 81? Also, is 0 a term of this AP? Give reasons for your answer.
Solution
Given:
We know,
For tₙ = −81:
So, −81 is the 35th term.
To check whether 0 is a term:
Therefore, 0 is the 8th term.
Final result:
−81 is the 35th term, and 0 is the 8th term.
Answer:−81 is the 35th term, and 0 is the 8th term.
Question 3
↑ Jump to NavigatorFind the nth term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.
Solution
Given:
Using:
we get,
Therefore:
Recursive rule:
The first term is:
Each term is obtained by subtracting 3 from the previous term:
tₙ = tₙ₋₁ − 3, for n ≥ 2
Answer:nth term is tₙ = 14 − 3n; Recursive rule: t₁ = 11, tₙ = tₙ₋₁ − 3 for n ≥ 2.
Question 4
↑ Jump to NavigatorAn AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.
💡Hint from textbook:
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.
We know,
Given:
and the last term (50th term):
Subtracting (1) from (2):
(a + 49d) − (a + 2d) = 106 − 12
Substituting d = 2 in (1):
For the 29th term:
Therefore:
Answer:t₂₉ = 64
Question 5
↑ Jump to NavigatorHow 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, …, 99
This is an AP with:
Using:
we get,
So, there are 30 two-digit numbers divisible by 3.
Their sum is:
we get,
Therefore:
Answer:Number of two-digit numbers = 30; Sum S₃₀ = 1665
Question 6
↑ Jump to NavigatorHarish 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
The yearly salaries form an AP:
Here:
We want the salary to reach ₹7,00,000.
Using:
we get,
Therefore, ₹7,00,000 is the 11th year's salary.
Since the starting salary is the first year's salary:
Number of years = 11 − 1 = 10
Therefore, his income reached that amount after 10 years.
Answer:After 10 years (as ₹7,00,000 is the salary for the 11th year, t₁₁).
Question 7
↑ Jump to NavigatorA 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
The number of marbles in the rows is:
1, 2, 3, …, 25
This is an AP with:
Total marbles:
we get,
Therefore:
325 marbles
Answer:Total marbles used S₂₅ = 325
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