Ch 8 EOC Class 9 Maths Ganita Manjari | 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 End of Chapter (EOC) Exercises. Designed to reflect how top-scoring students write their solutions and reasoning in school examinations, this comprehensive review guide covers advanced arithmetic progressions, geometric progressions, word problems, fractal sequences, and number patterns entirely.
🔑 Key Formulas & Concepts Covered:
- Arithmetic Progression (AP): \(t_n = a + (n-1)d\), Sum \(S_n = \frac{n}{2}[2a + (n-1)d]\).
- Geometric Progression (GP): \(t_n = ar^{n-1}\), where \(a\) is the first term and \(r\) is the common ratio.
- Recursive & Explicit Sequences: Understanding relationship formulas and pattern growth.
NCERT Class 9 Maths Ganita Manjari Chapter 8 End of Chapter Solutions
Question 1
Find the \(31^{\text{st}}\) term of an AP whose \(11^{\text{th}}\) term is 38 and \(16^{\text{th}}\) term is 73.
Solution:
Given \(t_{11} = 38 \implies a + 10d = 38\).
Given \(t_{16} = 73 \implies a + 15d = 73\).
Subtracting the first equation from the second:
\(5d = 35 \implies d = 7\).
Substitute \(d = 7\) into \(a + 10d = 38\):
\(a + 10(7) = 38 \implies a + 70 = 38 \implies a = -32\).
Now, find the \(31^{\text{st}}\) term (\(t_{31}\)):
\(t_{31} = a + 30d = -32 + 30(7) = -32 + 210 = \mathbf{178}\).
Answer: The \(31^{\text{st}}\) term is \(\mathbf{178}\).
Question 2
Determine the AP whose third term is 16 and whose \(7^{\text{th}}\) term exceeds the \(5^{\text{th}}\) term by 12.
Solution:
Given \(t_3 = 16 \implies a + 2d = 16\).
Given \(t_7 - t_5 = 12 \implies (a + 6d) - (a + 4d) = 12 \implies 2d = 12 \implies d = 6\).
Substitute \(d = 6\) into \(a + 2d = 16\):
\(a + 2(6) = 16 \implies a + 12 = 16 \implies a = 4\).
Constructing the AP: \(a, a+d, a+2d, \dots \implies 4, 10, 16, 22, \dots\)
Answer: The AP is \(\mathbf{4, 10, 16, 22, \dots}\).
Question 3
How many three-digit numbers are divisible by 7?
(Hint: All three-digit numbers divisible by 7 form an AP. Find the smallest and largest such three-digit numbers.)
Solution:
The first three-digit number divisible by 7 is 105, and the last is 994.
This forms an AP: \(105, 112, 119, \dots, 994\).
Here, \(a = 105\), \(d = 7\), and \(l = 994\).
Using \(t_n = a + (n-1)d\):
\(994 = 105 + (n-1)7 \implies 889 = 7(n-1) \implies n - 1 = 127 \implies n = 128\).
Answer: There are \(\mathbf{128}\) three-digit numbers divisible by 7.
Question 4
How many multiples of 4 lie between 10 and 250?
(Hint: All multiples of 4 form an AP. Find the smallest and largest multiples of 4 between 10 and 250.)
Solution:
The first multiple of 4 after 10 is 12, and the last multiple before 250 is 248.
This forms an AP: \(12, 16, 20, \dots, 248\).
Here, \(a = 12\), \(d = 4\), and \(l = 248\).
\(248 = 12 + (n-1)4 \implies 236 = 4(n-1) \implies n - 1 = 59 \implies n = 60\).
Answer: There are \(\mathbf{60}\) multiples of 4 between 10 and 250.
Question 5
Find a GP for which the sum of the first two terms is \(-4\) and the fifth term is 4 times the third term.
Solution:
Let the first term be \(a\) and common ratio be \(r\).
Given sum of first two terms \(= -4 \implies a + ar = -4 \implies a(1 + r) = -4\).
Given fifth term is 4 times the third term \(\implies ar^4 = 4ar^2 \implies r^2 = 4 \implies r = 2\) or \(r = -2\).
Case 1: When \(r = 2\)
\(a(1 + 2) = -4 \implies 3a = -4 \implies a = -\frac{4}{3}\).
GP: \(-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \dots\)
Case 2: When \(r = -2\)
\(a(1 - 2) = -4 \implies -a = -4 \implies a = 4\).
GP: \(4, -8, 16, -32, \dots\)
Answer: The possible GPs are \(\mathbf{-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \dots}\) or \(\mathbf{4, -8, 16, -32, \dots}\).
Question 6
Find all possible ways of expressing 100 as the sum of consecutive natural numbers.
Solution:
Let \(k\) consecutive natural numbers start from \(x\). Their sum is:
\(S = kx + \frac{k(k-1)}{2} = 100 \implies 2kx + k(k-1) = 200 \implies k(2x + k - 1) = 200\).
Here, \(k\) and \((2x + k - 1)\) are factors of 200 of opposite parity.
Factoring 200 into factor pairs where one is odd and one is even:
* If \(k = 1\) (odd), then \(2x + 1 - 1 = 200 \implies 2x = 200 \implies x = 100\) (Sum = \(\{100\}\)).
* If \(k = 8\) (even), then \(2x + 8 - 1 = 25 \implies 2x + 7 = 25 \implies 2x = 18 \implies x = 9\) (Sum = \(\{9, 10, 11, 12, 13, 14, 15, 16\}\)).
Answer: There are 2 ways: \(\mathbf{100}\) (single number) or \(\mathbf{9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100}\).
Question 7
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the \(2^{\text{nd}}\) hour, \(4^{\text{th}}\) hour and \(n^{\text{th}}\) hour?
Solution:
Initial bacteria (\(t = 0\)) \(= 30\).
Since the number doubles every hour, bacteria form a GP with \(a = 30\) and \(r = 2\).
* End of \(2^{\text{nd}}\) hour: \(30 \times 2^2 = 30 \times 4 = \mathbf{120}\).
* End of \(4^{\text{th}}\) hour: \(30 \times 2^4 = 30 \times 16 = \mathbf{480}\).
* End of \(n^{\text{th}}\) hour: \(\mathbf{30 \times 2^n}\).
Question 8
The sum of the \(4^{\text{th}}\) and \(8^{\text{th}}\) terms of an AP is 24 and the sum of the \(6^{\text{th}}\) and \(10^{\text{th}}\) terms is 44. Find the first three terms of the AP.
Solution:
\(t_4 + t_8 = 24 \implies (a + 3d) + (a + 7d) = 24 \implies 2a + 10d = 24 \implies a + 5d = 12\).
\(t_6 + t_{10} = 44 \implies (a + 5d) + (a + 9d) = 44 \implies 2a + 14d = 44 \implies a + 7d = 22\).
Subtracting the first equation from the second:
\(2d = 10 \implies d = 5\).
Substitute \(d = 5\) into \(a + 5d = 12\):
\(a + 5(5) = 12 \implies a + 25 = 12 \implies a = -13\).
First three terms: \(a = -13\), \(a+d = -8\), \(a+2d = -3\).
Answer: The first three terms are \(\mathbf{-13, -8, -3}\).
Question 9
Find the smallest value of \(n\) such that the sum of the first \(n\) natural numbers is greater than 1,000.
Solution:
Sum of first \(n\) natural numbers is \(S_n = \frac{n(n+1)}{2}\).
We need \(S_n > 1000 \implies \frac{n(n+1)}{2} > 1000 \implies n(n+1) > 2000\).
Testing integer values: Since \(\sqrt{2000} \approx 44.72\):
* For \(n = 44\): \(44 \times 45 = 1980\) (not greater than 2000).
* For \(n = 45\): \(45 \times 46 = 2070 > 2000\).
Answer: The smallest value of \(n\) is \(\mathbf{45}\).
Question 10
Which term of the GP: \(2, 8, 32, \dots\) is 131072? Write the explicit formula as well as the recursive formula for the \(n^{\text{th}}\) term.
Solution:
Given GP: \(2, 8, 32, \dots \implies a = 2\), common ratio \(r = 4\).
* Explicit formula: \(t_n = ar^{n-1} = \mathbf{2 \cdot 4^{n-1}}\).
* Recursive formula: \(t_1 = 2\) and \(t_{n+1} = 4t_n\).
Find which term is 131072:
\(2 \cdot 4^{n-1} = 131072 \implies 4^{n-1} = 65536 = 4^8 \implies n - 1 = 8 \implies n = 9\).
Answer: 131072 is the \(\mathbf{9^{\text{th}}}\) term.
Question 11
The sum of the first three terms of a GP is \(\frac{13}{12}\) and their product is \(-1\). Find the common ratio and the terms.
Solution:
Let the three terms be \(\frac{a}{r}, a, ar\).
Product \(= \frac{a}{r} \cdot a \cdot ar = a^3 = -1 \implies a = -1\).
Sum \(= \frac{-1}{r} - 1 - r = \frac{13}{12} \implies \frac{1}{r} + r = -\frac{25}{12} \implies 12r^2 + 25r + 12 = 0\).
Solving the quadratic equation: \((3r + 4)(4r + 3) = 0 \implies r = -\frac{3}{4}\) or \(r = -\frac{4}{3}\).
* If \(r = -\frac{3}{4}\), the terms are \(\mathbf{\frac{4}{3}, -1, -\frac{3}{4}}\).
* If \(r = -\frac{4}{3}\), the terms are \(\mathbf{\frac{3}{4}, -1, -\frac{4}{3}}\).
Answer: Common ratio is \(-\frac{3}{4}\) or \(-\frac{4}{3}\), and terms are \(\mathbf{\frac{4}{3}, -1, -\frac{3}{4}}\) (or vice versa).
Question 12
If the \(4^{\text{th}}\), \(10^{\text{th}}\) and \(16^{\text{th}}\) terms of a GP are \(x\), \(y\) and \(z\) respectively, prove that \(x, y, z\) are in GP.
Solution:
Let the first term be \(a\) and common ratio be \(r\).
\(x = t_4 = ar^3\)
\(y = t_{10} = ar^9\)
\(z = t_{16} = ar^{15}\)
Consider \(y^2 = (ar^9)^2 = a^2 r^{18}\).
Consider \(x \cdot z = (ar^3)(ar^{15}) = a^2 r^{18}\).
Since \(y^2 = xz\), the terms \(x, y, z\) satisfy the condition for a Geometric Progression.
Hence proved.
Question 13
The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP.
Solution:
Let terms be \(\frac{a}{r}, a, ar\).
Sum: \(\frac{a}{r} + a + ar = 26 \implies a\left(\frac{1}{r} + 1 + r\right) = 26\).
Sum of squares: \(\left(\frac{a}{r}\right)^2 + a^2 + (ar)^2 = 364 \implies a^2\left(\frac{1}{r^2} + 1 + r^2\right) = 364\).
Solving these simultaneous equations yields \(a = 6\) and \(r = 3\) (or \(r = \frac{1}{3}\)).
Answer: The terms of the GP are \(\mathbf{2, 6, 18}\) (or \(\mathbf{18, 6, 2}\)).
Question 14
Suppose \(P_1 = 1, P_2 = 2\) and for \(n > 2\), \(P_n = P_1 + P_2 + \dots + P_{n-1} + 1\). Find the values of \(P_1, P_2, \dots, P_8\). Can you find a simpler recursive formula for \(P_n\)? Can you give an explicit formula?
Solution:
* \(P_1 = 1\)
* \(P_2 = 2\)
* \(P_3 = 1 + 2 + 1 = 4\)
* \(P_4 = 1 + 2 + 4 + 1 = 8\)
* \(P_5 = 16, P_6 = 32, P_7 = 64, P_8 = 128\).
Simpler recursive formula: \(P_n = 2P_{n-1}\) for \(n \ge 2\) with \(P_1 = 1\).
Explicit formula: \(P_n = \mathbf{2^{n-1}}\) for \(n \ge 1\).
Question 15
Suppose \(W_1 = 1, W_2 = 2\) and for \(n > 2\), \(W_n = W_1 + W_2 + \dots + W_{n-2} + 2\). Find the values of \(W_1, W_2, \dots, W_8\). Do you recognise this sequence?
Solution:
* \(W_1 = 1\)
* \(W_2 = 2\)
* \(W_3 = 1 + 2 = 3\)
* \(W_4 = 1 + 2 + 2 = 5\)
* \(W_5 = 1 + 2 + 3 + 2 = 8\)
* \(W_6 = 13\), \(W_7 = 21\), \(W_8 = 34\).
Answer: This sequence is the famous Fibonacci sequence.
Frequently Asked Questions (FAQs)
Q1: How do you solve problems involving sums of consecutive natural numbers?
A: Set up the sum formula using AP or factoring divisors of twice the target sum.
Q2: What is the relationship between the terms of a geometric progression?
A: The ratio of any term to its preceding term is constant, denoted as the common ratio \(r\).
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