Additional Questions - Chapter 2 - Numbers and Sequences - 10th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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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 $12576=4052 \times 3+420$.
Since the remainder $420 \neq 0$, we apply the division lemma to 4052 $4052=420 \times 9+272$.
We consider the new divisor 420 and the new remainder 272 and apply the division lemma to get $420=272 \times 1+148,148 \neq 0$.
$\therefore$ Again by division lemma
$272=148 \times 1+124$, here $124 \neq 0$.
$\therefore$ Again by division lemma
$148=124 \times 1+24$, Here $24 \neq 0$.
$\therefore$ Again by division lemma
$124=24 \times 5+4$, Here $4 \neq 0$.
$\therefore$ Again by division lemma
$24=4 \times 6+0$.
The remainder has now become zero. So our procedure stops. Since the divisor at this stage is 4 .
$\therefore$ The HCF of 12576 and 4052 is 4 .

Question 2.
If the HCF of 65 and 117 is in the form $(65 \mathrm{~m}-117)$ then find the value of $\mathrm{m}$.
Answer:
By Euclid's algorithm $117>65$
$117=65 \times 1+52$
$52=13 \times 4 \times 0$
$65=52 \times 1+13$
H.C.F. of 65 and 117 is 13
$65 \mathrm{~m}-117=13$
$65 \mathrm{~m}=130$

$\mathrm{m}=\frac{130}{65}=2$
The value of $m=2$
 

Question 3.
Find the LCM and HCF of 6 and 20 by the prime factorisation method.
Solution:
We have $6=2^{1} \times 3^{1}$ and
$20=2 \times 2 \times 5=2^{2} \times 5^{1}$
You can find $\operatorname{HCF}(6,20)=2$ and $\operatorname{LCM}(6,20)=2 \times 2 \times 3 \times 5=60$. 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.
$\operatorname{LCM}(6,20)=2^{2} \times 3^{1} \times 5^{1}=60$.
= 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 $\mathrm{p}$ and $\mathrm{q}(\mathrm{q} \neq 0)$ are co-prime (no common factor other than 1)
Hence, $\sqrt{3}=\frac{\underline{p}}{q}$
$\sqrt{3} q=p$
Squaring both side
$\begin{aligned}
(\sqrt{3} q)^{2} &=p^{2} \\
3 q^{2} &=p^{2} \\
q^{2} &=\frac{p^{2}}{3}
\end{aligned}$
Hence, 3 divides $\mathrm{p}^{2}$
So 3 divides $p$ also ................ (1)
Hence we can say

$\frac{p}{3}=\mathrm{c}$ where $\mathrm{c}$ is some integer
$\mathrm{p}=3 \mathrm{c}$
Now we know that
$3 q^{2}=p^{2}$
$\begin{aligned}
&\text { Putting }=3 \mathrm{c} \\
&3 \mathrm{q}^{2}=(3 \mathrm{c})^{2} \\
&3 \mathrm{q}^{2}=9 \mathrm{c}^{2} \\
&\mathrm{q}^{2}=\frac{1}{3} \times 9 \mathrm{c}^{2} \\
&\mathrm{q}^{2}=3 \mathrm{c}^{2} \\
&\frac{q^{2}}{3}=\mathrm{C}^{2}
\end{aligned}$
Hence 3 divides $q^{2}$
So, 3 divides $q$ also
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) $4,10,16,22, \ldots$
(ii) $1,-1,-3,-5, \ldots$
(iii) $-2,2,-2,2,-2, \ldots$
(iv) $1,1,1,2,2,2,3,3,3, \ldots$
Solution:
(i) $4,10,16,22, \ldots \ldots$.
We have $a_{2}-a_{1}=10-4=6$

$\begin{aligned}
&\mathrm{a}_{3}-\mathrm{a}_{2}=16-10=6 \\
&\mathrm{a}_{4}-\mathrm{a}_{3}=22-16=6
\end{aligned}$
$\therefore$ It is an A.P. with common difference 6 .
$\therefore$ The next two terms are, $\underline{28}, \underline{34}$
$\begin{aligned}
&\text { (ii) } 1,-1,-3,-5 \\
&t_{2}-t_{1}=-1-1=-2 \\
&t_{3}-t_{2}=-3-(-1)=-2 \\
&t_{4}-t_{3}=-5-(-3)=-2
\end{aligned}$
The given list of numbers form an A.P with the common difference $-2$. The next two terms are $(-5+(-2))=-7,-7+(-2)=-9$.
$\begin{aligned}
&\text { (iii) }-2,2,-2,2,-2 \\
&t_{2}-t_{1}=2-(-2)=4 \\
&t_{3}-t_{2}=-2-2=-4 \\
&t_{4}-t_{3}=2-(-2)=4
\end{aligned}$
It is not an A.P.
(iv) $1,1,1,2,2,2,3,3,3$
$\begin{aligned}
&t_{2}-t_{1}=1-1=0 \\
&t_{3}-t_{2}=1-1=0 \\
&t_{4}-t_{3}=2-1=1
\end{aligned}$
Here $t_{2}-t_{1} \neq t_{3}-t_{2}$
$\therefore$ It is not an A.P.

Question $6 .$
Find $\mathrm{n}$ so that the $\mathrm{n}^{\text {th }}$ terms of the following two A.P.'s are the same.
$1,7,13,19, \ldots$ and $100,95,90, \ldots$
Answer:
The given A.P. is $1,7,13,19, \ldots$.
$\begin{aligned}
&\mathrm{a}=1, \mathrm{~d}=7-1=6 \\
&\mathrm{t}_{\mathrm{n} 1}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d} \\
&\mathrm{t}_{\mathrm{n} 1}=1+(\mathrm{n}-1) 6 \\
&=1+6 \mathrm{n}-6=6 \mathrm{n}-5 \ldots(1)
\end{aligned}$
The given A.P. is $100,95,90, \ldots$
$\mathrm{a}=100, \mathrm{~d}=95-100=-5$ $\mathrm{tn}_{2}=100+(\mathrm{n}-1)(-5)$ $=100-5 \mathrm{n}+5$ $=105-5 \mathrm{n} \ldots \ldots(2)$ Given that, $\mathrm{t}_{\mathrm{n} 1}=\mathrm{t}_{\mathrm{n} 2}$ $6 \mathrm{n}-5=105-5 \mathrm{n}$ $6 \mathrm{n}+5 \mathrm{n}=105+5$ $11 \mathrm{n}=110$ $\mathrm{n}=10$ $\therefore 10^{\text {th }}$ 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?
Answer:
The number of rose plants in the $1 \mathrm{st}$, 2nd, 3rd,. rows are
$23,21,19, \ldots \ldots . \ldots .5$
It forms an A.P.
Let the number of rows in the flower bed be $\mathrm{n}$.
Then $\mathrm{a}=23, \mathrm{~d}=21-23=-2,1=5$.
As, $a_{n}=a+(n-1)$ d i.e. $t_{n}=a+(n-1) d$
We have $5=23+(n-1)(-2)$
i.e. $-18=(n-1)(-2)$

$\mathrm{n}-10$
$\therefore$ There are 10 rows in the flower bed.
 

Question $8 .$
Find the sum of the first 30 terms of an A.P. whose $n^{\text {th }}$ term is $3+2 n$.
Answer:
Given,
$\begin{aligned}
&t_{n}=3+2 n \\
&t_{1}=3+2(1)=3+2=5 \\
&t_{2}=3+2(2)=3+4=7 \\
&t_{3}=3+2(3)=3+6=9
\end{aligned}$
Here $\mathrm{a}=5, \mathrm{~d}=7-5=2, \mathrm{n}=30$
$\mathrm{S}_{\mathrm{n}}=\frac{n}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$ $\mathrm{S}_{30}=\frac{30}{2}[10+29(2)]$ $=15[10+58]=15 \times 68=1020$ $\therefore$ 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 $\mathrm{a}=24, \mathrm{~d}=21-24=-3, \mathrm{~S}_{\mathrm{n}}=78$. We need to find $\mathrm{n}$.
We know that,
$\begin{aligned}
&\mathrm{S}_{\mathrm{n}}=\frac{n}{2}(2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}) \\
&78=\frac{\pi}{2}(48+13(-3)) \\
&78=\frac{n}{2}(5 \mathrm{l}-3 \mathrm{n})
\end{aligned}$
or $3 n^{2}-5 \ln +156=0$
$\begin{aligned}
&\mathrm{n}^{2}-17 \mathrm{n}+52=0 \\
&(\mathrm{n}-4)(\mathrm{n}-13)=0 \\
&\mathrm{n}=4 \text { or } 13
\end{aligned}$
The number of terms are 4 or 13 .

Question $10 .$
The sum of first $n$ terms of a certain series is given as $3 n^{2}-2 n$. Show that the series is an arithmetic series.
Solution:
Given, $S_{n}=3 n^{2}-2 n$
$S_{1}=3(1)^{2}-2(1)$
$=3-2=1$
ie; $\mathrm{t}_{1}=1\left(\therefore \mathrm{S}_{1}=\mathrm{t}_{1}\right)$
$S_{2}=3(2)^{2}-2(2)=12-4=8$
ie; $\mathrm{t}_{1}+\mathrm{t}_{2}=8\left(\therefore \mathrm{S}_{2}=\mathrm{t}_{1}+\mathrm{t}_{2}\right)$
$\therefore \mathrm{t}_{2}=8-\mathrm{l}=7$
$=3-2=1$
ie; $t_{1}=1\left(\therefore S_{1}=t_{1}\right)$
$S_{2}=3(2)^{2}-2(2)=12-4=8$
ie $^{;} t_{1}+t_{2}=8\left(\therefore S_{2}=t_{1}+t_{2}\right)$
$\therefore t_{2}=8-1=7$
$S_{3}=3(3)^{2}-2(3)=27-6=21$
$t_{1}+t_{2}+t_{3}=21\left(\therefore S_{3}=t_{1}+t_{2}+t_{3}\right)$
$8+t_{3}=21\left(\right.$ Substitute $\left.t_{1}+t_{2}=8\right)$
$t_{3}=21-8 \Rightarrow t_{3}=13$
$S_{3}=3(3)^{2}-2(3)=27-6=21$
$t_{1}+t_{2}+t_{3}=21\left(\therefore S_{3}=t_{1}+t_{2}+t_{3}\right)$
$8+t_{3}=21$ (Substitute $t_{1}+t_{2}=8$ )
$\mathrm{t}_{3}=21-8 \Rightarrow \mathrm{t}_{3}=13$
$\therefore$ The series is $1,7,13, \ldots \ldots \ldots \ldots .$ and this series is an A.P. with common difference 6 .