10th Maths - Book Back Answers - Chapter 2 Exercise 2.4 - English Medium Guides

 

SSLC / 10th - Maths - Book Back Answers - Chapter 2 Excercise 2.4 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 2 Excercise 2.4: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 2 Excercise 2.4 – Maths from the Tamil Nadu State Board 10th Standard Maths textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.

    We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.

    By going through this material, you’ll gain a strong understanding of Chapter 2 Excercise 2.4 along with the corresponding book back questions and answers (PDF format).

Question Types Covered:

• 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following 
• 2 Mark Questions: Answer briefly 
• 3, 4, and 5 Mark Questions: Answer in detail

All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.

All the best, Class 10 students! Prepare well and aim for top scores. Thank you!

Chapter 2 Numbers and Sequences Ex 2.4

1. Find the next three terms of the following sequence.
(i) 8, 24, 72,…
(ii) 5, 1, -3, …
(iii) $\frac{\mathbf{1}}{\mathbf{4}}\mathbf{,}\frac{\mathbf{2}}{\mathbf{9}}\mathbf{,}\frac{\mathbf{3}}{\mathbf{16}}$

Answer Key: 

(i) 216, 648, 1944 (This sequence is multiple of 3)
Next three terms are 216, 648, 1944

(ii) Next three terms are -7, -11, -15
(adding -4 with each term)

(iii) Next three terms are $\frac{4}{25},\frac{5}{36}$ and $\frac{6}{49}$
[using $\frac{\mathrm{n}}{(\mathrm{n}+1{)}^2}$]

2. Find the first four terms of the sequences whose nth terms are given by
(i) a_n = n³ – 2
(ii) a_n = (-1)ⁿ⁺¹ n(n+1)
(iii) a_n = 2n² – 6

Answer Key:

t_n = a_n = n³ -2

(i) a₁ = 1³ – 2 = 1 – 2 – 1
a₂ = 2³ – 2 = 8 – 2 = 6
a₃ = 3³ – 2 = 27 – 2 = 25
a₄ = 4³ – 2 = 64 – 2 = 62
∴ The first four terms are -1, 6, 25, 62, ……….

(ii) a_n = (-1)ⁿ⁺¹ n(n + 1)
a₁ = (-1)¹⁺¹ (1) (1 +1)
= (-1)² (1) (2) = 2
a₂ = (-1)²⁺¹ (2) (2 + 1)
= (-1)³ (2) (3)= -6
a₃ = (-1)³⁺¹ (3) (3 + 1)
= (-1)⁴ (3) (4) = 12
a₄ = (-1)⁴⁺¹ (4) (4 + 1)
= (-1)⁵ (4) (5) = -20
∴ The first four terms are 2, -6, 12, -20,…

 

(iii) a_n = 2n² – 6
a₁ = 2(1)² – 6 = 2 – 6 = -4
a₂ = 2(2)² – 6 = 8 – 6 = 2
a₃ = 2(3)² – 6 = 18 – 6 = 12
a₄ = 2(4)² – 6 = 32 – 6 = 26
∴ The first four terms are -4, 2, 12, 26, …

 

3. Find the n^th term of the following sequences

Answer Key:

(i) 2, 5, 10, 17, ……
(1² + 1);(2² + 1),(3² + 1),(4² + 1)….
n^th term is n² + 1
a_n = n² + 1

 

$\mathbf{(}$$\mathbf{i}$$\mathbf{i}\mathbf{)}\mathbf{ }\mathbf{0}\mathbf{,}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$ 

($\frac{\mathbf{1}\mathbf{-}\mathbf{1}}{\mathbf{1}}$),($\frac{\mathbf{2}\mathbf{-}\mathbf{1}}{\mathbf{2}}$),($\frac{3-1}{3}$)....

n^th term is $\frac{n-1}{n}$
a_n = $\frac{n-1}{n}$

 

(iii) 3,8,13,18,…….
[5(1) -2], [5(2) – 2], [5(3) – 2], [5(4) – 2] ….
The n^th term is 5n – 2
a_n = 5n – 2

 

4. Find the indicated terms of the sequences whose n^th terms are given by

Answer Key:

(i)${\mathbf{a}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{5}\mathbf{n}}{\mathbf{n}\mathbf{+}\mathbf{2}}$ ; a₆ and a₁₃
$a_6=\frac{5\left(6\right)}{6+2}=\frac{30}{8}=\frac{15}{4}$

$a_{13}=\frac{5\left(13\right)}{13+2}=\frac{5\times 13}{15}=\frac{13}{3}$

a₆ = $\frac{15}{4}$
a₁₃ = $\frac{13}{3}$

(ii) a_n = – (n² – 4); a₄ and a₁₁
a_n = -(n² – 4)
a₄ = -(4² – 4)
= – (16 – 4)
= -12
a₁₁ = -(11² – 4)
= – (121 – 4)
= – 117
a₄ = -12 and a₁₁ = -117

 

5. Find a₈ and a₁₅ whose n^th term is a_n 

Answer Key:

${\mathrm{a}}_{\mathrm{n}}=\frac{{\mathrm{n}}^2-1}{\mathrm{n}+3}$

$a_8=\frac{8^2-1}{8+3}=\frac{64-1}{11}=\frac{63}{11}$

${\mathrm{a}}_{\mathrm{n}}=\frac{{\mathrm{n}}^2}{2\mathrm{n}+1}$

${\mathrm{a}}_{15}=\frac{{15}^2}{2\left(15\right)+1}=\frac{225}{31}$

$a_8=\frac{63}{11}$

${\mathrm{a}}_{15}=\frac{225}{31}$ 

 

6. If a₁ = 1, a₂ = 1 and a_n = 2a_n-1 + a_n-2, n > 3, n ∈ N, then find the first six terms of the sequence.

Answer Key:

a₁ = 1, a₂ = 1, a_n = 2a_n-1 + a_n-2
a₃ = 2a_(3-1) + a_(3-2)
= 2a₂ + a₁
= 2 × 1 + 1 = 3
a₄ = 2a_(4-1) + a_(4-2)
= 2a₃ + a₂
= 2 × 3 + 1 = 7
a₅ = 2a_(5-1) + a_(5-2)
= 2a₄ + a₃
= 2 × 7 + 3 = 17
a₆ = 2a_(6-1) + a_(6-2)
= 2a₅ + a₄
= 2 × 17 + 7
= 34 + 7
= 41
∴ The first six terms of the sequence are 1, 1, 3, 7, 17, 41 ………..