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## Samacheer Kalvi 10th Class Maths Chapter 2 Additional Questions

**Question 1. **Use Euclid’s algorithm to find the HCF of 4052 and 12756.

**Solution**

Since 12576 > 4052 we apply the division lemma to 12576 and 4052, to get HCF

12576 = 4052 × 3 + 420.

Since the remainder 420 ≠ 0, we apply the division lemma to 4052

4052 = 420 × 9 + 272.

We consider the new divisor 420 and the new remainder 272 and apply the division lemma to get

420 = 272 × 1 + 148, 148 ≠ 0.

∴ Again by division lemma

272 = 148 × 1 + 124, here 124 ≠ 0.

∴ Again by division lemma

148 = 124 × 1 + 24, Here 24 ≠ 0.

∴ Again by division lemma

124 = 24 × 5 + 4, Here 4 ≠ 0.

∴ Again by division lemma

24 = 4 × 6 + 0.

The remainder has now become zero. So our procedure stops. Since the divisor at this stage is 4.

∴ The HCF of 12576 and 4052 is 4.

**Question 2. **If the HCF of 65 and 117 is in the form (65m – 117) then find the value of m.

**Solution**

By Euclid’s algorithm 117 > 65

117 = 65 × 1 + 52

52 = 13 × 4 × 0

65 = 52 × 1 + 13

H.C.F. of 65 and 117 is 13

65m – 117 = 13

65 m = 130

m = 130/65 = 2

The value of m = 2

The smallest number to appear on both lists is 60, so 60 is the least common of 15 and 20.

**Question 3. **Find the LCM and HCF of 6 and 20 by the prime factorisation method.

**Solution**

We have 6 = 2^{1} × 3^{1} and

20 = 2 × 2 × 5 = 2^{2} × 5^{1}

You can find HCF (6, 20) = 2 and LCM (6, 20) = 2 × 2 × 3 × 5 = 60. As done in your earlier classes. Note that HCF (6, 20) = 2^{1} = product of the smallest power of each common prime factor in the numbers.

LCM (6, 20) = 2^{2} × 3^{1} × 5^{1} = 60.

= Product of the greatest power of each prime factor, involved in the numbers.

Common Multiples of 16: 16, 32, 48, 64, 80,… Hence the Least common multiple of 12 and 16 is 48. The LCM of 12 and 16 is 48.

**Question 4. **Prove that √3 is irrational.

**Solution**

**Question 5.** Which of the following list of numbers form an AP? If they form an AP, write the next two terms:

(i) 4, 10, 16, 22, …

(ii) 1, -1,-3, -5,…

(iii) -2, 2, -2, 2, -2, …

(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3,…

**Solution**

(i) 4, 10, 16, 22, …….

We have a_{2} – a_{1} = 10 – 4 = 6

a_{3} – a_{2} = 16 – 10 = 6

a_{4} – a_{3} = 22 – 16 = 6

∴ It is an A.P. with common difference 6.

∴ The next two terms are, 28, 34

(ii) 1, -1, -3, -5

t_{2} – t_{1} = -1 – 1 = -2

t_{3} – t_{2} = -3 – (-1) = -2

t_{4} – t_{3} = -5 – (-3) = -2

The given list of numbers form an A.P with the common difference -2.

The next two terms are (-5 + (-2)) = -7, -7 + (-2) = -9.

(iii) -2, 2,-2, 2,-2

t_{2} – t_{1} = 2-(-2) = 4

t_{3} – t_{2} = -2 -2 = -4

t_{4} – t_{3} = 2 – (-2) = 4

It is not an A.P.

(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3

t_{2} – t_{1} = 1 – 1 = 0

t_{3} – t_{2} = 1 – 1 = 0

t_{4} – t_{3} = 2 – 1 = 1

Here t_{2} – t_{1} ≠ t_{3} – t_{2}

∴ It is not an A.P.

**Question 6. **Find n so that the n^{th} terms of the following two A.P.’s are the same.

1, 7,13,19,… and 100, 95,90,…

**Solution**

The given A.P. is 1, 7, 13, 19,….

a = 1, d = 7 – 1 = 6

t_{n1} = a + (n – 1)d

t_{n1} = 1 + (n – 1) 6

= 1 + 6n – 6 = 6n – 5 … (1)

The given A.P. is 100, 95, 90,….

a = 100, d = 95 – 100 = – 5

tn_{2} = 100 + (n – 1) (-5)

= 100 – 5n + 5

= 105 – 5n …..(2)

Given that, t_{n1} = t_{n2}

6n – 5 = 105 – 5n

6n + 5n = 105 + 5

11 n = 110

n = 10

∴ 10^{th} term are same for both the A.P’s.

**Question 7. **In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 is the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?

**Solution**

The number of rose plants in the 1st, 2nd, 3rd,… rows are

23, 21, 19,………….. 5

It forms an A.P.

Let the number of rows in the flower bed be n.

Then a = 23, d = 21 – 23 = -2, l = 5.

As, a_{n} = a + (n – 1)d i.e. t_{n} = a + (n – 1)d

We have 5 = 23 + (n – 1)(-2)

i.e. -18 = (n – 1)(-2)

n = 10

∴ There are 10 rows in the flower bed.

**Question 8. **Find the sum of the first 30 terms of an A.P. whose n^{th} term is 3 + 2n.

**Solution**

Given,

t_{n} = 3 + 2n

t_{1} = 3 + 2 (1) = 3 + 2 = 5

t_{2} = 3 + 2 (2) = 3 + 4 = 7

t_{3} = 3 + 2 (3) = 3 + 6 = 9

Here a = 5,d = 7 – 5 = 2, n = 30

S_{n} = n2 [2a + (n – 1)d]

S_{30} = 302 [10 + 29(2)]

= 15 [10 + 58] = 15 × 68 = 1020

∴ Sum of first 30 terms = 1020

**Question 9. **How many terms of the AP: 24, 21, 18, . must be taken so that their sum is 78?

**Solution**

Here a = 24, d = 21 – 24 = -3, S_{n} = 78. We need to find n.

We know that,

S_{n} = n2 (2a + (n – 1)d)

78 = n2 (48 + 13(-3))

78 = n2 (51 – 3n)

or 3n^{2} – 51n + 156 = 0

n^{2} – 17n + 52 = 0

(n – 4) (n – 13) = 0

n = 4 or 13

The number of terms are 4 or 13.

**Question 10. **The sum of first n terms of a certain series is given as 3n^{2} – 2n. Show that the series is an arithmetic series.

**Solution**

Given, S_{n} = 3n^{2} – 2n

S_{1} = 3 (1)^{2} – 2(1)

= 3 – 2 = 1

ie; t_{1} = 1 (∴ S_{1} = t_{1})

S_{2} = 3(2)^{2} – 2(2) = 12 – 4 = 8

ie; t_{1} + t_{2} = 8 (∴ S_{2} = t_{1} + t_{2})

∴ t_{2} = 8 – 1 = 7

S_{3} = 3(3)^{2} – 2(3) = 27 – 6 = 21

t_{1} + t_{2} + t_{3} = 21 (∴ S_{3} = t_{1} + t_{2} + t_{3})

8 + t_{3} = 21 (Substitute t_{1} + t_{2} = 8)

t_{3} = 21 – 8 ⇒ t_{3} = 13

∴ The series is 1,7,13, …………. and this series is an A.P. with common difference 6.

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