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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Additional Questions
Question 1.
Use EuclidтАЩs 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 ├Ч 3 + 420.
Since the remainder 420 тЙа 0, we apply the division lemma to 4052
4052 = 420 ├Ч 9 + 272.
We consider the new divisor 420 and the new remainder 272 and apply the division lemma to get
420 = 272 ├Ч 1 + 148, 148 тЙа 0.
тИ┤ Again by division lemma
272 = 148 ├Ч 1 + 124, here 124 тЙа 0.
тИ┤ Again by division lemma
148 = 124 ├Ч 1 + 24, Here 24 тЙа 0.
тИ┤ Again by division lemma
124 = 24 ├Ч 5 + 4, Here 4 тЙа 0.
тИ┤ Again by division lemma
24 = 4 ├Ч 6 + 0.
The remainder has now become zero. So our procedure stops. Since the divisor at this stage is 4.
тИ┤ 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.
Answer:
By Euclid’s algorithm 117 > 65
117 = 65 ├Ч 1 + 52
52 = 13 ├Ч 4 ├Ч 0
65 = 52 ├Ч 1 + 13
H.C.F. of 65 and 117 is 13
65m – 117 = 13
65 m = 130
m = \(\frac { 130 }{ 65 } \) = 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 ├Ч 31 and
20 = 2 ├Ч 2 ├Ч 5 = 22 ├Ч 51
You can find HCF (6, 20) = 2 and LCM (6, 20) = 2 ├Ч 2 ├Ч 3 ├Ч 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 ├Ч 31 ├Ч 51 = 60.
= Product of the greatest power of each prime factor, involved in the numbers.
Common Multiples of 16: 16, 32, 48, 64, 80,тАж Hence the Least common multiple of 12 and 16 is 48. The LCM of 12 and 16 is 48.
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 тЙа 0) are co-prime (no common factor other than 1)
Hence, 3 divides p2
So 3 divides p also …………….. (1)
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 = \(\frac { 1 }{ 3 } \) ├Ч 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 – a1 = 10 – 4 = 6
a3 – a2 = 16 – 10 = 6
a4 – a3 = 22 – 16 = 6
тИ┤ It is an A.P. with common difference 6.
тИ┤ 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 тЙа t3 – t2
тИ┤ It is not an A.P.
Question 6.
Find n so that the nth terms of the following two A.P.тАЩs are the same.
1, 7,13,19,… and 100, 95,90,…
Answer:
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
тИ┤ 10th term are same for both the A.PтАЩs.
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
тИ┤ 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.
Answer:
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 = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S30 = \(\frac { 30 }{ 2 } \) [10 + 29(2)]
= 15 [10 + 58] = 15 ├Ч 68 = 1020
тИ┤ 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 = \(\frac { n }{ 2 } \) (2a + (n – 1)d)
78 = \(\frac { n}{ 2 } \) (48 + 13(-3))
78 = \(\frac { n}{ 2 } \) (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 (тИ┤ S1 = t1)
S2 = 3(2)2 – 2(2) = 12 – 4 = 8
ie; t1 + t2 = 8 (тИ┤ S2 = t1 + t2)
тИ┤ t2 = 8 – 1 = 7
S3 = 3(3)2┬а– 2(3) = 27 – 6 = 21
t1 + t2 + t3 = 21 (тИ┤ S3 = t1 + t2 + t3)
8 + t3 = 21 (Substitute t1 + t2 = 8)
t3 = 21 – 8 тЗТ t3┬а= 13
тИ┤ The series is 1,7,13, …………. and this series is an A.P. with common difference 6.
