10th Maths - Book Back Answers - Chapter 2 Exercise 2.6 - English Medium Guides

 

SSLC / 10th - Maths - Book Back Answers - Chapter 2 Exercise 2.6 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 2 Exercise 2.6: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 2 Exercise 2.6 – 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 2.6 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 2.6

1. Find the sum of the following
(i) 3, 7, 11,… up to 40 terms.
Answer Key:
3,7,11,… up to 40 terms
First term (a) = 3
Common difference (d) = 7 – 3 = 4
Number of terms (n) = 40
S_n = $\frac{\mathbf{n}}{\mathbf{2}}$[2a + (n – 1) d]
S₄₀ = $\frac{\mathbf{40}}{\mathbf{2}}$[6 + 39 × 4] = 20 [6 + 156]
= 20 × 162
S₄₀ = 3240

 

(ii) 102,97, 92,… up to 27 terms.
Answer Key:
Here a = 102, d = 97 – 102 = -5
n = 27
S_n = $\frac{\mathbf{n}}{\mathbf{2}}$ [2a + (n – 1)d]
S₂₇ = $\frac{\mathbf{27}}{\mathbf{2}}$[2(102) + 26(-5)]
=$\frac{\mathbf{27}}{\mathbf{2}}$ [204 – 130]
= $\frac{\mathbf{27}}{\mathbf{2}}$× 74
= 27 × 37 = 999
S₂₇ = 999

 

(iii) 6 + 13 + 20 + …. + 97
Answer Key:
Here a = 6, d = 13 – 6 = 7, l = 97
n =$\frac{\mathbf{l}\mathbf{-}\mathbf{a}}{\mathbf{d}}$+ 1
= $\frac{\mathbf{97}\mathbf{-}\mathbf{6}}{\mathbf{7}}$+ 1
$\mathbf{=}\frac{\mathbf{91}}{\mathbf{7}}\mathbf{+}\mathbf{1}$
13 + 1 = 14
S_n =$\frac{\mathbf{n}}{\mathbf{2}}$$\frac{\mathbf{(a+1)}}{}$
S_n =$\frac{\mathbf{14}}{\mathbf{2}}$(a+1)
S_n =$\frac{\mathbf{14}}{\mathbf{2}}$$\frac{\mathbf{(6+97)}}{}$
= 7 × 103
S_n = 721

 

2. How many consecutive odd integers beginning with 5 will sum to 480?
Answer Key:
First term (a) = 5
Common difference (d) = 2
(consecutive odd integer)
S_n = 480

$\frac{\mathbf{n}}{\mathbf{2}}$[2a + (n-1) d] = 480

$\frac{\mathbf{n}}{\mathbf{2}}$[10 + (n-1)2] = 480

n + 24 = 0 or n – 20 = 0
n = -24 or n = 20
[number of terms cannot be negative]
∴ Number of consecutive odd integers is 20

 

3. Find the sum of first 28 terms of an A.P. whose n^th term is 4n – 3.
Answer Key:
n = 28
t_n = 4n – 3
t₁ = 4 × 1 – 3 = 1
t₂ = 4 × 2 – 3 = 5
t₂₈ = 4 × 28 – 3
= 112 – 3 = 109
∴ a = 1, d = t₂ – t₁ = 5 – 1 = 4
l = 109.
S_n =$\frac{\mathbf{n}}{\mathbf{2}}$(2a+(n – 1)d)
S₂₈ =$\frac{\mathbf{28}}{\mathbf{2}}$(2 × 1 + 27 × 4)
= 14(2 + 108)
= 14 × 110
= 1540

 

4. The sum of first n terms of a certain series is given as 2n² – 3n . Show that the series is an A.P.
Answer Key:
Let tn be nth term of an A.P.
t_n = Sn – S_n-1
= 2n² – 3n – [2(n – 1)² – 3(n – 1)]
= 2n² – 3n – [2(n² – 2n + 1) – 3n + 3]
= 2n² – 3n – [2n² – 4n + 2 – 3n + 3]
= 2n² – 3n – [2n² – 7n + 5]
= 2n² – 3n – 2n² + 7n – 5
t_n = 4n – 5
t₁ = 4(1) – 5 = 4 – 5 = -1
t₂ = 4(2) -5 = 8 – 5 = 3
t₃ = 4(3) – 5 = 12 – 5 = 7
t₄ = 4(4) – 5 = 16 – 5 = 11
The A.P. is -1, 3, 7, 11,….
The common difference is 4
∴ The series is an A.P.

 

5. The 104th term and 4^th term of an A.P are 125 and 0. Find the sum of first 35 terms?
Answer Key:
104th term of an A.P = 125
t₁₀₄ = 125
[t_n = a + (n – 1) d]
a + 103d = 125 …..(1)
4^th term = 0
t₄ = 0
a + 3d = 0 …..(2)

Sum of 35 terms = 612.5

 

6. Find the sum of ail odd positive integers less than 450.
Answer Key:
Sum of odd positive integer less than 450
1 + 3 + 5 + …. 449
Here a = 1, d = 3 – 1 = 2,l = 449

Aliter: Sum of all the positive odd intergers
= n²
= 225 × 225
= 50625
Sum of the odd integers less than 450
= 50625

 

7. Find the sum of all natural numbers between 602 and 902 which are not divisible by 4?
Answer Key:
First find the sum of all the natural’s number between 602 and 902
Here a = 603, d = 1, l = 901

Find the sum of all the numbers between 602 and 902 which are divisible by 4.
Here a = 604; l = 900; d = 4

Sum of the numbers which are not divisible
by 4 = S_n1 – S_n2
= 224848 – 56400
= 168448
Sum of the numbers = 168448

 

8. Raghu wish to buy a laptop. He can buy it by paying ₹40,000 cash or by giving it in 10 installments as ₹4800 in the first month, ₹4750 in the second month, ₹4700 in the third month and so on. If he pays the money in this fashion, find
(i) total amount paid in 10 installments.
(ii) how much extra amount that he has to pay than the cost?
Answer Key:
 4800 + 4750 + 4700 + … 10 terms
Here a = 4800
(i) d = t₂ – t₁ = 4750 – 4800 = -50
n = 10
S_n =$\frac{\mathbf{n}}{\mathbf{2}}$(2a + (n – 1)d)
S₁₀ =$\frac{\mathbf{10}}{\mathbf{2}}$(2 × 4800 + 9 × -50)
= 5 (9600 – 450)
= 5 × 9150 = 45750
Total amount paid in 10 installments = ₹ 45750.
(ii) The extra amount he pays in installments
= ₹ 45750 – ₹ 40,000
= ₹ 5750

 

9. A man repays a loan of ₹ 65,000 by paying ₹ 400 in the first month and then increasing the payment by ₹ 300 every month. How long will it take for him to clear the loan?
Answer Key:
(i) Total loan amount = ₹ 65,000
S_n = 65,000
First month payment (a) = 400
Every month increasing ₹ 300
d = 300
S_n =$\frac{\mathbf{n}}{\mathbf{2}}$[2a + (n-1)d]
65000 =$\frac{\mathbf{n}}{\mathbf{2}}$[2(400) + (n – 1)300]
130000 = n [800 + 300n – 300]
= n [500 + 300n]
13000 = 500n + 300n²
Dividing by (100)

 Number of installments will not be negative
∴ Time taken to pay the loan is 20 months.

10. A brick staircase has a total of 30 steps. The bottom step requires 100 bricks. Each successive step requires two bricks less than the previous step.
(i) How many bricks are required for the top most step?
(ii) How many bricks are required to build the stair case?
Answer Key:
100 + 98 + 96 + 94 + … 30 steps.
Here
a = 100
d = -2
n = 30
∴ S_n =$\frac{\mathbf{n}}{\mathbf{2}}$(2a + (n – 1)d)
S₃₀ =$\frac{\mathbf{30}}{\mathbf{2}}$(2 × 100 + 29 × -2)
= 15(200 – 58)
= 15 × 142
= 2130
t₃₀ = a + (n – 1)d
= 100 + 29 × -2
= 100 – 58
= 42
(i) No. of bricks required for the top step are 42.
(ii) No. of bricks required to build the stair case are 2130.

 

11. If S₁, S₂ , S₃, ….S_m are the sums of n terms of m A.P.,s whose first terms are 1,2, 3…… m and whose common differences are 1,3,5,…. (2m – 1) respectively, then show that (S₁ + S₂ + S₃ + ……. + S_m) =$\frac{\mathbf{1}}{\mathbf{2}}$mn(mn + 1)
Answer Key:
First terms of an A.P are 1, 2, 3,…. m
The common difference are 1, 3, 5,…. (2m – 1)

By adding (1) (2) (3) we get
S₁ + S₂ + S₃ + …… + S_m =$\frac{\mathbf{n}}{\mathbf{2}}$(n + 1) +$\frac{\mathbf{n}}{\mathbf{2}}$(3n + 1) +$\frac{\mathbf{n}}{\mathbf{2}}$(5n + 1) + ….. +$\frac{\mathbf{n}}{\mathbf{2}}$[n(2m – 1 + 1)]
=$\frac{\mathbf{n}}{\mathbf{2}}$[n + 1 + 3n + 1 + 5n + 1 ……. + n (2m – 1) + m)]
=$\frac{\mathbf{n}}{\mathbf{2}}$[n + 3n + 5n + ……. n(2m – 1) + m]
=$\frac{\mathbf{n}}{\mathbf{2}}$[n (1 + 3 + 5 + ……(2m – 1)) + m
=$\frac{\mathbf{n}}{\mathbf{2}}$[n($\frac{\mathbf{m}}{\mathbf{2}}$)(2m) + m]
=$\frac{\mathbf{n}}{\mathbf{2}}$[nm² + m]
S₁ + S₂ + S₃ + ……….. + S_m =$\frac{\mathbf{m}\mathbf{n}}{\mathbf{2}}$[mn + 1]
Hint:
1 + 3 + 5 + ……. + 2m – 1
S_n =$\frac{\mathbf{n}}{\mathbf{2}}$(a + l)
=$\frac{\mathbf{m}}{\mathbf{2}}$(1 + 2m -1)
=$\frac{\mathbf{m}}{\mathbf{2}}$(2m)

 

12. Find the sum

Answer Key: