Ex 8.2 Class 9 Maths Ganita Manjari Solutions | Sid Classes
Welcome to Sid Classes, your trusted destination for comprehensive academic resources! In this detailed study guide, we present complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.2. Designed to reflect how top-scoring students write their solutions and reasoning in school examinations, this walkthrough makes arithmetic progressions, general terms, common differences, sums, and word problem applications entirely clear.
🔑 Key Formulas & Concepts Used in this Exercise:
- Arithmetic Progression (AP): A sequence where each term after the first is obtained by adding a fixed number (common difference \(d\)) to the preceding term.
- General Term (\(n^{\text{th}}\) term): \(t_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference.
- Sum of \(n\) terms of an AP: \(S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}(a + l)\), where \(l\) is the last term.
NCERT Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.2 Solutions
Question 1
Find the \(10^{\text{th}}\) and \(26^{\text{th}}\) terms of the AP: \(3, 8, 13, 18, \dots\)
Solution:
Given AP: \(3, 8, 13, 18, \dots\)
First term \(a = 3\), common difference \(d = 8 - 3 = 5\).
* For \(10^{\text{th}}\) term (\(t_{10}\)):
\(t_{10} = a + (10 - 1)d = 3 + 9(5) = 3 + 45 = \mathbf{48}\).
* For \(26^{\text{th}}\) term (\(t_{26}\)):
\(t_{26} = a + (26 - 1)d = 3 + 25(5) = 3 + 125 = \mathbf{128}\).
Answer: The \(10^{\text{th}}\) term is \(48\) and the \(26^{\text{th}}\) term is \(128\).
Question 2
Which term of the AP : \(21, 18, 15, \dots\) is \(-81\)? Also, is \(0\) a term of this AP? Give reasons for your answer.
Solution:
Given AP: \(21, 18, 15, \dots\)
First term \(a = 21\), common difference \(d = 18 - 21 = -3\).
(i) Which term is \(-81\)?
Let \(t_n = -81\). Using \(t_n = a + (n-1)d\):
\(21 + (n-1)(-3) = -81 \implies -3(n-1) = -81 - 21 \implies -3(n-1) = -102 \implies n - 1 = 34 \implies n = 35\).
Since \(n = 35\) is a positive integer, \(-81\) is the \(\mathbf{35^{\text{th}}}\) term.
(ii) Is \(0\) a term of this AP?
Let \(t_n = 0\):
\(21 + (n-1)(-3) = 0 \implies -3(n-1) = -21 \implies n - 1 = 7 \implies n = 8\).
Since \(n = 8\) is a positive integer, \(\mathbf{0}\) is a term (specifically the \(8^{\text{th}}\) term) of this AP.
Question 3
Find the \(n^{\text{th}}\) term of the AP: \(11, 8, 5, 2, \dots\) Write the recursive rule for this AP.
Solution:
Given AP: \(11, 8, 5, 2, \dots\)
First term \(a = 11\), common difference \(d = 8 - 11 = -3\).
\(n^{\text{th}}\) term formula:
\(t_n = a + (n-1)d = 11 + (n-1)(-3) = 11 - 3n + 3 = \mathbf{14 - 3n}\).
Recursive rule:
\(t_1 = 11\) and \(t_{n+1} = t_n - 3\) for \(n \ge 1\).
Question 4
An AP consists of 50 terms in which the \(3^{\text{rd}}\) term is 12 and the last term is 106. Find the \(29^{\text{th}}\) 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:
Given total terms \(n = 50\), so the last term is \(t_{50} = 106\).
Given \(3^{\text{rd}}\) term \(t_3 = 12 \implies a + 2d = 12\).
Last term \(t_{50} = 106 \implies a + 49d = 106\).
Subtracting the first equation from the second:
\((a + 49d) - (a + 2d) = 106 - 12 \implies 47d = 94 \implies d = \frac{94}{47} = \mathbf{2}\).
Substitute \(d = 2\) into \(a + 2d = 12\):
\(a + 2(2) = 12 \implies a + 4 = 12 \implies a = \mathbf{8}\).
Now, find the \(29^{\text{th}}\) term (\(t_{29}\)):
\(t_{29} = a + 28d = 8 + 28(2) = 8 + 56 = \mathbf{64}\).
Answer: The \(29^{\text{th}}\) term is \(\mathbf{64}\).
Question 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 form an AP: \(12, 15, 18, \dots, 99\).
Here, first term \(a = 12\), common difference \(d = 3\), and last term \(l = 99\).
(i) Find how many terms (\(n\)) there are:
\(t_n = a + (n-1)d \implies 99 = 12 + (n-1)3 \implies 87 = 3(n-1) \implies n - 1 = 29 \implies n = 30\).
There are \(\mathbf{30}\) two-digit numbers divisible by 3.
(ii) Find the sum of these numbers (\(S_{30}\)):
\(S_{30} = \frac{n}{2}(a + l) = \frac{30}{2}(12 + 99) = 15 \times 111 = \mathbf{1665}\).
Answer: There are \(30\) numbers, and their sum is \(\mathbf{1665}\).
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?
Solution:
The annual salaries form an AP: \(5,00,000, 5,20,000, 5,40,000, \dots, 7,00,000\).
Here, initial salary \(a = 5,00,000\), annual increment \(d = 20,000\), and final salary \(t_n = 7,00,000\).
Using \(t_n = a + (n-1)d\):
\(5,00,000 + (n-1)20,000 = 7,00,000\)
\((n-1)20,000 = 2,00,000 \implies n - 1 = \frac{2,00,000}{20,000} = 10 \implies n = 11\).
Since \(n = 11\) represents the \(11^{\text{th}}\) year of work, his income reached ₹7,00,000 after \(\mathbf{10}\) years of increment (in the \(11^{\text{th}}\) year).
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?
Solution:
The number of marbles in each row forms an AP: \(1, 2, 3, \dots\) up to \(25\) terms.
Here, first term \(a = 1\), common difference \(d = 1\), and number of rows \(n = 25\).
Total marbles \(= S_{25} = \frac{n}{2}[2a + (n-1)d]\):
\(S_{25} = \frac{25}{2}[2(1) + (25-1)(1)] = \frac{25}{2}[2 + 24] = \frac{25}{2} \times 26 = 25 \times 13 = \mathbf{325}\).
Answer: The child uses \(\mathbf{325}\) marbles in all.
Frequently Asked Questions (FAQs)
Q1: What is the formula for the \(n^{\text{th}}\) term of an AP?
A: \(t_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference.
Q2: How do you find the sum of an AP when the first and last terms are known?
A: Use the formula \(S_n = \frac{n}{2}(a + l)\).
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