Samacheer Kalvi 10th Maths Book Solutions Chapter 2 | Numbers and Sequences Answers For Tamil Nadu Board

Samacheer Kalvi 10th Maths Book Solutions Chapter 2 – Numbers and Sequences are available here. The Samacheer Kalvi 10th Maths book answers of Chapter 2, available at BYJU’S, contain step by step explanations designed by our mathematics experts. All these important questions are based on the new pattern prescribed by the Tamil Nadu board. Students can also get the solutions of other chapters on samacheer kalvi 10th maths solutions.

Chapter 2 of the Samacheer Kalvi 10th Maths guide will help the students to solve problems related to Euclid’s division lemma, Euclid’s division algorithm, the highest common factor of three numbers, fundamental theorem of arithmetic and its significance, modular arithmetic, congruence modulo, connecting Euclid’s division lemma and modular arithmetic, modulo operations, sequences, arithmetic progression, series, geometric progression, special series.

Samacheer Kalvi 10th Maths Chapter 2: Numbers and Sequences Book Exercise 2.1 Questions and Solutions

Question 1: If d is the Highest Common Factor of 32 and 60, find x and y satisfying d = 32x + 60y.

Solution:

60 > 32

60 = 32(1) + 28 —-(1)

32 = 28(1) + 4 —-(2)

28 = 4(7) + 0 —-(3)

H.C.F of 60 and 32 is 4.

d = 4

d = 32x + 60y

4 = 32 x + 60 y

From (2), solve for 4

4 = 32 – 28(1) —(4)

From (1), solve for 28

28 = 60 – 32(1) —(5)

Putting the value of 28 from (5) into (4),

4 = 32 – (60 – 32(1)) ⋅ 1

4 = 32 – 60 ⋅ 1 + 32 (1) ⋅ 1

4 = 32 + 32 (1) – 60 ⋅ 1

4 = 32(1 + 1) – 60 ⋅ 1

4 = 32(2) + 60 (-1)

Hence x = 2 and y = -1.

Samacheer Kalvi 10th Maths Chapter 2: Numbers and Sequences Book Exercise 2.2 Questions and Solutions

Question 1: Find the greatest number consisting of 6 digits which is exactly divisible by 24,15,36.

Solution:

Find the LCM of 24,15 and 36 which is 360.

The greatest 6-digit number is 999999.

Divide 999999 number by LCM of 24,15 and 36,

If 999999 / 360, then the remainder is 279.

Now 999999 – 279 = 999720

On checking for the numbers,

999720 / 24 = 41655

999720 / 15 = 66648

999720 / 36 = 27770

999720 is the greatest 6-digit number that is divisible by 24,15 and 36.

Samacheer Kalvi 10th Maths Chapter 2: Numbers and Sequences Book Exercise 2.3 Questions and Solutions

Question 1: Solve 5x ≡ 4 (mod 6)

Solution:

5x – 4 = 6n

n = (5x – 4) / 6

n = [(6 – 1)x – (6 – 2)] / 6

n = 6 (-1x + 2) / 6

n = (-1x + 2)

If x = 2, then n = -2 + 2 = 0

If x = 8, then n = -8 + 2 = -6

If x = 14, then n = -14 + 2 = -12

The values of x are 2, 8, 14,…….

Samacheer Kalvi 10th Maths Chapter 2: Numbers and Sequences Book Exercise 2.4 Questions and Solutions

Question 1: If a1 = 1, a2 = 1 and an = 2an – 1 + an – 2 n ≥ 3, n ∈ N, then find the first six terms of the sequence.

Solution:

First and second terms are 1.

an = 2an – 1 + an – 2

n = 3

an = 2an – 1 + an – 2

a3 = 2a3-1 + a3-2

a3 = 2a2 + a1

a3 = 2(1) + 1

a3 = 3

n = 4

an = 2an – 1 + an – 2

a4 = 2a4-1 + a4-2

a4 = 2a3 + a2

a4 = 2(3) + 1

a4 = 7

n = 5

an = 2an – 1 + an – 2

a5 = 2a5-1 + a5-2

a5 = 2a4 + a3

a5 = 2(7) + 3

a5 = 17

n = 6

an = 2an – 1 + an – 2

a6 = 2a6-1 + a6-2

a6 = 2a5 + a4

a5 = 2(17) + 7

a5 = 41

The first 6 terms are 1, 1, 3, 7, 17, 41.

Samacheer Kalvi 10th Maths Chapter 2: Numbers and Sequences Book Exercise 2.5 Questions and Solutions

Question 1: 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?

Solution:

Number of seats in 1st row = 20

Number of seats in 2nd row = 20 + 2 = 22

Number of seats in 3rd row = 22 + 2 = 24

Total number of rows in theatre (n) = 30

a = 20, d = 2 and n = 30

Number of seats in the last row be tn or “l”

n = [(l – a) / d] + 1

30 = [(l – 20) / 2] + 1

(30 – 1) = (l – 20) / 2

29 (2) = l – 20

l – 20 = 58

l = 58 + 20

l = 78

So, the number of seats in the last row is 78.

Samacheer Kalvi 10th Maths Chapter 2: Numbers and Sequences Book Exercise 2.6 Questions and Solutions

Question 1: Find the sum of all odd positive integers less than 450.

Solution:

Write the odd positive numbers less than 450,

1, 3, 5, 7, ………..447, 449

1 + 3 + 5 + 7 +…………. + 447 + 449

a = 1, d = 3 – 1 = 2, l = 449

Sn = (n / 2) [a + l]

n = [(l – a) / d] + 1

n = [(449 – 1) / 2] + 1

= (448 / 2) + 1

n = 225

Sn = (225 / 2) [1 + 449]

= (225 / 2) 450

S225 = 50625

Samacheer Kalvi 10th Maths Chapter 2: Numbers and Sequences Book Exercise 2.7 Questions and Solutions

Question 1: A man joined a company as Assistant Manager. The company gave him a starting salary of ₹60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years?

Solution:

Starting salary = 60,000

Every year there exists a 5% increase in the annual salary.

Second year salary = 60000 + 5% of 60000

= 60000(1 + 5%)

Third year salary = 60000 + 5% of 60000(1 + 5%)

= 60000(1 + 5%)(1 + 5%)

= 60000(1 + 5%)2

By continuing in this way, the salary after 5 years will be

= 60000(1 + 5%)5

= 60000(105/100)5

= 60000(1.05)5

= 76577

Samacheer Kalvi 10th Maths Chapter 2: Numbers and Sequences Book Exercise 2.8 Questions and Solutions

Question 1: Find the sum of the geometric series 3 + 6 + 12 +…………+ 1536

Solution:

tn = 1536

a = 3, r = 6 / 3 = 2

arn-1 = 1536

3(2)n-1 = 1536

2n-1 = 1536 / 3

2n-1 = 512

2n-1 = 29

n – 1 = 9

n = 10

To find the sum of 10 terms.

Sn = a(rn – 1) / (r – 1)

S10 = 3(210 – 1) / (2 – 1)

= 3(1024 – 1)

= 3(1023)

S10 = 3069

Samacheer Kalvi 10th Maths Chapter 2: Numbers and Sequences Book Exercise 2.9 Questions and Solutions

Question 1: Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm,…, 24 cm. How much area can be decorated with these colour papers?

Solution:

Required area

102 + 112 + 122 + ………….+ 242

= (12 + 22 + 32 + ………+ 242) – (12 + 22 + 32 + ……….+ 92)

= [24(24 + 1) (2(24) + 1) / 6] – [9(9 + 1)(2(9) + 1) / 6]

= [24(25)(49) / 6] – [9(10)(19) / 6]

= 4900 – 285

= 4615 cm2

Question 2: Find the sum of the series (23 − 1) + (43 −33) + (63 −53) + ……. to

(i) n terms

(ii) 8 terms

Solution:

(23 − 1) + (43 −33) + (63 −53) + …….

First numbers are even and second numbers are odd.

(i) n terms

The general term of the given series = (2n)3 − (2n – 1)3

= 8n3 – [(2n)3 – 3(2n)2 (1) + 3(2n)(1) – 13]

= 8n3 – [8n3 – 12n2 + 6n – 1]

= 8n3 – 8n3 + 12n2 – 6n + 1

= 12n2 – 6n + 1

= [12n(n +1)(2n + 1) / 6] – 6[n(n + 1) / 2] + n

= 2n(n + 1)(2n + 1) – 3n(n + 1) + n

= 2n(2n2 + n + 2n + 1) – 3n2 – 3n + n

= 4n3 + 6n2 + 2n – 3n2 – 3n + n

= 4n3 + 3n2

Hence, the sum of n terms 4n3 + 3n2

(ii) 8 terms

Sn = 4n3 + 3n2

n = 8

= 4(8)3 + 3(8)2

= 2048 + 192

= 2240

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