Samacheer Kalvi 10th Class Maths Chapter 2 Additional Questions

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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 = 21 × 31 and
20 = 2 × 2 × 5 = 22 × 51
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) = 21 = product of the smallest power of each common prime factor in the numbers.
LCM (6, 20) = 22 × 31 × 51 = 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 a2 – a1 = 10 – 4 = 6
a3 – a2 = 16 – 10 = 6
a4 – a3 = 22 – 16 = 6
∴ It is an A.P. with common difference 6.
∴ The next two terms are, 28, 34

(ii) 1, -1, -3, -5
t2 – t1 = -1 – 1 = -2
t3 – t2 = -3 – (-1) = -2
t4 – t3 = -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
t2 – t1 = 2-(-2) = 4
t3 – t2 = -2 -2 = -4
t4 – t3 = 2 – (-2) = 4
It is not an A.P.

(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3
t2 – t1 = 1 – 1 = 0
t3 – t2 = 1 – 1 = 0
t4 – t3 = 2 – 1 = 1
Here t2 – t1 ≠ t3 – t2
∴ It is not an A.P.

Question 6. Find n so that the nth 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
tn1 = a + (n – 1)d
tn1 = 1 + (n – 1) 6
= 1 + 6n – 6 = 6n – 5 … (1)
The given A.P. is 100, 95, 90,….
a = 100, d = 95 – 100 = – 5
tn2 = 100 + (n – 1) (-5)
= 100 – 5n + 5
= 105 – 5n …..(2)
Given that, tn1 = tn2
6n – 5 = 105 – 5n
6n + 5n = 105 + 5
11 n = 110
n = 10
∴ 10th 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, an = a + (n – 1)d i.e. tn = 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 nth term is 3 + 2n.

Solution

Given,
tn = 3 + 2n
t1 = 3 + 2 (1) = 3 + 2 = 5
t2 = 3 + 2 (2) = 3 + 4 = 7
t3 = 3 + 2 (3) = 3 + 6 = 9
Here a = 5,d = 7 – 5 = 2, n = 30
Sn = n2 [2a + (n – 1)d]
S30 = 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, Sn = 78. We need to find n.
We know that,
Sn = n2 (2a + (n – 1)d)
78 = n2 (48 + 13(-3))
78 = n2 (51 – 3n)
or 3n2 – 51n + 156 = 0
n2 – 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 3n2 – 2n. Show that the series is an arithmetic series.

Solution

Given, Sn = 3n2 – 2n
S1 = 3 (1)2 – 2(1)
= 3 – 2 = 1
ie; t1 = 1 (∴ S1 = t1)
S2 = 3(2)2 – 2(2) = 12 – 4 = 8
ie; t1 + t2 = 8 (∴ S2 = t1 + t2)
∴ t2 = 8 – 1 = 7
S3 = 3(3)2 – 2(3) = 27 – 6 = 21
t1 + t2 + t3 = 21 (∴ S3 = t1 + t2 + t3)
8 + t3 = 21 (Substitute t1 + t2 = 8)
t3 = 21 – 8 ⇒ t3 = 13
∴ The series is 1,7,13, …………. and this series is an A.P. with common difference 6.

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