10th Maths - Book Back Answers - Chapter 2 Exercise 2.5 - English Medium Guides

 

SSLC / 10th - Maths - Book Back Answers - Chapter 2 Exercise 2.5 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 2 Exercise 2.5: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 2 Exercise 2.5 – 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 Exercise 2.5 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 2.5

1. Check whether the following sequences are in A.P.?

(i) a – 3, a – 5, a – 7,…
Answer Key:

a – 3, a – 5, a – 7…….
t₂ – t₁ = a – 5 – (a – 3)
= a – 5 – a + 3
= -2
t₃ – t₂ = a – 7 – (a – 5)
= a – 7 – a + 5
= -2
t₂ – t₁ = t₃ – t₂
(common difference is same)
The sequence is in A.P.

 

(ii) $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{,}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{,}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$
Answer Key:
t₂ – t₁ = $\frac{1}{3}-\frac{1}{2}=\frac{2-3}{6}=\frac{-1}{6}$
t₃ – t₂ = $\frac{1}{4}-\frac{1}{3}=\frac{3-4}{12}=\frac{-1}{12}$
t₂ – t₁ ≠ t₃ – t₂
The sequence is not in A.P.

 

(iii) 9, 13, 17, 21, 25,…
Answer Key:
t₂ – t₁ = 13 – 9 = 4
t₃ – t₂ = 17 – 13 = 4
t₄ – t₃ = 21 – 17 = 4
t₅ – t₄ = 25 – 21 = 4
Common difference are equal
∴ The sequence is in A.P.

 

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

Answer Key: 
t₂ – t₁ = 0 – ($-\frac{1}{3}$)
$0+\frac{1}{3}=\frac{1}{3}$
t₃ – t₂ = $\frac{1}{3}$ – 0 =$\frac{1}{3}$
t₂ – t₁ = t₃ – t₂
The sequence is in A.P.

 

(v) 1,-1, 1,-1, 1, -1, …

Answer Key:
t₂ – t₁ = -1 – 1 = -2
t₃ – t₂ = 1 – (-1) = 1 + 1 = 2
t₄ – t₃ = -1-(1) = – 1 – 1 = – 2
t₅ – t₄ = 1 – (-1) = 1 + 1 = 2
Common difference are not equal

∴ The sequence is not an A.P.

 

2. First term a and common difference d are given below. Find the corresponding A.P. ?
(i) a = 5 ,d = 6
Answer Key:
Here a = 5,d = 6
The general form of the A.P is a, a + d, a + 2d, a + 3d….
The A.P. 5, 11, 17, 23 ….

 

(ii) a = 7, d = -5
Answer Key:

The general form of the A.P is a, a + d,
a + 2d, a + 3d… .
The A.P. 7, 2, -3, -8 ….

 

(iii) a =$\frac{3}{4}$, d = $\frac{1}{2}$
Answer Key:
The general form of the A.P is a, a + d, a + 2d, a + 3d….
$\frac{3}{4},\frac{3}{4}+\frac{1}{2},\frac{3}{4}$ + 2($\frac{1}{2}$), $\frac{3}{4}$+ 3 ($\frac{1}{2}$)
The A.P. $\frac{3}{4},\frac{5}{4},\frac{7}{4}.....$

 

3. Find the first term and common difference of the Arithmetic Progressions whose n^th terms are given below
(i) t_n = -3 + 2n
(ii) t_n = 4 – 7n

(i) t_n = -3 + 2n 

Answer Key: 
 a = t₁ = -3 + 2(1) = -3 + 2 = -1
d = t₂ – t₁
Here t₂ = -3 + 2(2) = -3 + 4 = 1
∴ d = t₂ – t₁ = 1 – (-1) = 2  

 

(ii) t_n = 4 – 7n

Answer Key: 
a = t₁ = 4 – 7(1) = 4 – 7 = -3
d = t₂ – t₁
Here t₂ = 4 – 7(2) = 4 – 14 – 10
∴ d = t₂ – t₁ = 10 – (-3) = -7

 

4. Find the 19^th term of an A.P. -11, -15, -19,…
Answer Key:

First term (a) = -11
Common difference (d) = -15 -(-11)
= -15 + 11 = -4
n = 19
t_n = a + (n – 1) d
t_n = -11 + 18(-4)
= -11 – 72
t₁₉ = -83
19^th term of an A.P. is – 83

 

5. Which term of an A.P. 16, 11, 6,1, ……….. is -54?
Answer Key:
 A.P = 16, 11,6, 1, ………..
It is given that
t_n = -54
a = 16, d = t₂ – t₁ = 11 – 16 = -5
∴ t_n = a + (n – 1)d
-54 = 16 + (n – 1) (-5)
-54 = 16 – 5n + 5
21 – 5n = -54
-5n = -54 -21
-5n = -75
n = $\frac{75}{5}$=15
∴ 15th term is -54.

 

6. Find the middle term(s) of an A.P. 9, 15, 21, 27, …, 183.
Answer Key:
First term (a) = 9
Last term (l) = 183
Common difference (d) = 15 – 9 = 6
$n=\frac{l-a}{d}+1$

$n=\frac{183-9}{6}+1$

$n=\frac{174}{6}+1$
= 29 + 1
= 30
middle term = 15^th term of
16^th term
t_n = a + (n – 1)d
t₁₅ = 9 + 14(6)
= 9 + 84 = 93
t₁₆ = 9 + 15(6)
= 9 + 90 = 99
The middle term is 93 or 99

 

7. If nine times ninth term is equal to the fifteen times fifteenth term, show that six times twenty fourth term is zero.
Answer Key:
 Nine times ninth term = Fifteen times fifteenth term
9t₉ = 15t₁₅
9(a + 8d) = 5(a + 14d)
9a + 72d = 15a + 210
15a + 210d – 9a – 72d = 0
⇒ 6a + 138 d = 0
⇒ 6(a + 23 d) = 0
⇒ 6(a + (24 – 1)d) = 0
⇒ 6t₂₄ = 0. Hence it is proved.

 

8. If 3 + k, 18 – k, 5k + 1 are in A.P. then find k?
Answer Key:
3 + k, 18 – k, 5k + 1 are in AP
∴ t₂ – t₁ = t₃ – t₂ (common difference is same)
18 – k – (3 + k) = 5k + 1 – (18 – k)
18 – k – 3 – k = 5k + 1 – 18 + k
15 – 2k = 6k – 17
32 = 8k
k = $\frac{32}{8}$= 4
The value of k = 4

 

9. Find x, y and z, given that the numbers x, 10, y, 24, z are in A.P.
Answer Key:
 A.P = x, 10, y, 24, z,…
d = t₂ – t₁ = 10 – x ………….. (1)
= t₃ – t₂ = y – 10 ………….. (2)
= t₄ – t₃ = 24 – y …………. (3)
= t₅ – t₄ = z – 24 ………….. (4)
(2) and (3)
⇒ y – 10 = 24 – y
2y = 24 + 10 = 34
y = $\frac{34}{2}$ = 17
(1) and (2)
⇒ 10 – x = y – 10
10 – x = 17 – 10 = 7
-x = 7 – 10
-x = -3 ⇒ x = 3
From (3) and (4)
24 – y = z – 24
24 – 17 = z – 24
7 = z – 24
∴ z = 7 + 24 = 31
∴ Solutions x = 3
y = 17
z = 31

 

10. In a theatre, there are 20 seats in the front row and 30 rows were allotted. Each successive row contains two additional seats than its front row. How many seats are there in the last row?
Answer Key:

Number of seats in the first row
(a) = 20
∴ t₁ = 20
Number of seats in the second row
(t₂) = 20 + 2
= 22
Number of seats in the third row
(t₃) = 22 + 2
= 24
Here a = 20 ; d = 2
Number of rows
(n) = 30
t_n = a + (n – 1)d
t₃₀ = 20 + 29(2)
= 20 + 58
t₃₀ = 78
Number of seats in the last row is 78

 

11. The sum of three consecutive terms that are in A.P. is 27 and their product is 288. Find the three terms.
Answer Key:
 Let the three consecutive terms be a – d, a, a + d
Their sum = a – d + a + a + d = 27
3a = 27
a =$\frac{27}{3}$= 9
Their product = (a – d)(a)(a + d) = 288
= 9(a² – d²) = 288
⇒ 9(9 – d²) = 288
⇒ 9(81 – d²) = 288
81 – d² = 32
-d² = 32 – 81
d² = 49
⇒ d = ± 7
∴ The three terms are if a = 9, d = 7
a – d, a , a + d = 9 – 7, 9 + 7
A.P. = 2, 9, 16
if a = 9, d = -7
A.P. = 9 – (-7), 9, 9 + (-7)
= 16, 9, 2

 

12. The ratio of 6^th and 8^th term of an A.P is 7:9. Find the ratio of 9^th term to 13^th term.
Answer Key:
Given : t₆ : t₈ = 7 : 9 (using t_n = a + (n – 1)d
a + 5d : a + 7d = 7 : 9
9 (a + 5 d) = 7 (a + 7d)
9a + 45 d = 7a + 49d
9a – 7a = 49d – 45d
2a = 4d
a = 2d
To find t₉ : t₁₃
t₉ : t₁₃ = a + 8d : a + 12d
= 2d + 8d : 2d + 12d
= 10d : 14d
= 5 : 7
∴ t₉ : t₁₃ = 5 : 7

 

13. In a winter season, the temperature of Ooty from Monday to Friday to be in A.P. The sum of temperatures from Monday to Wednesday is 0° C and the sum of the temperatures from Wednesday to Friday is 18° C. Find the temperature on each of the five days.
Answer Key:
Let the five days temperature be
(a – 2d), (a – d), a,(a + d) and (a + 2d)
Sum of first three days temperature = 0
a – 2d + a – d + a = 0
3a – 3d = 0
a – d = 0 …..(1)
Sum of the last three days temperature = 18°C
a + a + d + a + 2d = 18
3a + 3d = 18
(÷ by 3) ⇒ a + d = 6 ……(2)
By adding (1) and (2)
(1)+(2)⇒(a – d = 0)+(a + d = 6 )

(1)+(2)⇒2a=6 

a=$\frac{6}{2}$ =3
 Substitute to value of a = 3 in (2)
d = 3
The temperature in 5 days are
(3 – 6), (3 – 3), 3, (3 + 3) and (3 + 6)
-3°C, 0°C, 3°C, 6°C, 9°C

 

14. Priya earned ₹15,000 in the first month. Thereafter her salary increases by ₹1500 per year. Her expenses are ₹13,000 during the first year and the expenses increases by ₹900 per year. How long will it take her to save ₹20,000 per month.
Answer Key:

Duration of the year	Monthly salary	Monthly expenses	Monthly Saving
I year	15000	13000	2000
II year	16500	13900	2600
III year	18000	14800	3200

Monthly savings form an A.P.
2000, 2600, 3200 …..
a = 2000; d = 2600 – 2000 = 600
Given t_n = 20,000
t_n = a + (n – 1) d
20000 = 2000 + (n – 1) 600
20000 = 2000 + 600n – 600
= 1400 + 600n
20000 – 1400 = 600n
18600 = 600n
n = $\frac{18600}{600}$ = 31
He will take 31 years to save ₹ 20,000 per month