Tamilnadu State Board Class 10 Maths Solutions Chapter 2 Numbers and Sequences 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.
Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.
Solution:
Let us start with taking a, where a is a +ve odd integer.
We apply the division algorithm with ‘a’ and ‘b’ = 4.
Since 0 < r < 4, the possible remainders are 0,1,2,3.
That is, a can be 4q, or 4q + 1, or 4q + 2 or 4q + 3, where 1 is the quotient. However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2).
Any odd integer is of the form 4q + 1 or 4q + 3.
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.
Question 4.
Prove that \(\sqrt { 3 }\) is irrational.
Answer:
Let us assume the opposite, (1) \(\sqrt { 3 }\) is irrational.
Hence \(\sqrt { 3 }\) = \(\frac { p }{ q } \)
Where p and q(q ≠ 0) are co-prime (no common factor other than 1)
Hence, 3 divides p2
So 3 divides p also
Hence we can say
\(\frac { p }{ 3 } \) = c where c is some integer
p = 3c
Now we know that
3q2 = p2
Putting = 3c
3q2 = (3c)2
3q2 = 9c2
q2 = \(\frac { 1 }{ 3 } \) × 9c2
q2 = 3c2
\(\frac{q^{2}}{3}\) = C2
Hence 3 divides q2
So, 3 divides q also …..(2)
By (1) and (2) 3 divides both p and q
By contradiction \(\sqrt { 3 }\) is irrational.
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 – a2 = 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.
Determine the AP whose 3rd term is 5 and the 7th term is 9.
Solution:
We have
a3 = a + (3 – 1)d = a + 2d = 5 (1)
a7 = a + (7 – 1)d = a + 6d = 9 (2)
(1) – (2) ⇒ -4d = -4 ⇒ d = 1.
Sub, d = 1 in (1), we get
a + 2(1) = 5
a = 3
Hence the required A.P. is 3, 4, 5, 6, 7.
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?
Answer:
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.
If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.
Solution:
Here S14 = 1050
n = 14
a = 10
Sn = \(\frac { n }{ 2 } \) (2a + (n – 1)d)
1050 = \(\frac { 14 }{ 2 } \) (20 + 13d)
= 140 + 91d
910 = 91d
d = 10
a20 = 10 + (20 – 1) × 10
= 20
∴ 20th term = 200.
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,
The number of terms are 4 or 13.
Question 10.
Find the sum of first 24 terms of the list of numbers whose nth term is given by
an = 3 + 2n.
Solution:
an = 3 + 2n
a1 = 3 + 2 = 5
a2 = 3 + 2 × 2 = 7
a3 = 3 + 2 × 3 = 9
List of numbers becomes 5, 7, 9. 11,…
Here,7 – 5 = 9 – 7 = 11 – 9 = 2 and soon. So, it forms an A.P. with common difference d = 2.
To find S24, we have n = 24, a = 5, d = 2.
S24 = \(\frac { 24 }{ 2 } \) [2 × 5 + (24 – 1) × 2 ]
= 12 [10 + 46] = 672
So, sum of first 24 terms of the list of numbers is 672.