Unit Exercise 2 - Chapter 2 Numbers and Sequences 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Unit Exercise 2
Question 1.
Prove that $\mathrm{n}^{2}-\mathrm{n}$ divisible by 2 for every positive integer $\mathrm{n}$.
Answer:
To prove $\mathrm{n}^{2}-\mathrm{n}$ divisible by 2 for every positive integer $\mathrm{n}$.
We know that any positive integer is of the form $2 \mathrm{q}$ or $2 \mathrm{q}+1$, for some integer $\mathrm{q}$.
So, following cases arise:
Case I. When $n=2 q$.
In this case, we have
$\mathrm{n}^{2}-\mathrm{n}=(2 \mathrm{q})^{2}-2 \mathrm{q}=4 \mathrm{q}^{2}-2 \mathrm{q}=2 \mathrm{q}(2 \mathrm{q}-1)$
$\Rightarrow \mathrm{n}^{2}-\mathrm{n}=2 \mathrm{r}$ where $\mathrm{r}=\mathrm{q}(2 \mathrm{q}-1)$
$\Rightarrow \mathrm{n}^{2}-\mathrm{n}$ is divisible by 2 .
Case II. When $n=2 q+1$.
In this case, we have
$\begin{aligned}
&\mathrm{n}^{2}-\mathrm{n}=(2 \mathrm{q}+1)^{2}-(2 \mathrm{q}+1) \\
&=(2 \mathrm{q}+1)(2 \mathrm{q}+1-1)=(2 \mathrm{q}+1) 2 \mathrm{q} \\
&\Rightarrow \mathrm{n}^{2}-\mathrm{n}=2 \mathrm{r} \text { where } \mathrm{r}=\mathrm{q}(2 \mathrm{q}+1) \\
&\Rightarrow \mathrm{n}^{2}-\mathrm{n} \text { is divisible by } 2 .
\end{aligned}$
Hence $\mathrm{n}^{2}-\mathrm{n}$ is divisible by 2 for every positive integer $\mathrm{n}$.
 

Question $2 .$
A milk man has 175 litres of cow's milk and 105 litres of buffalow's milk. He wishes to sell the milk by filling the two types of milk in cans of equal capacity. Calculate the following (i) Capacity of a can
(ii) Number of cans of cow's milk
(iii) Number of cans of huffalow's milk
Answer:
Cow's milk $=175$ litres

Buffalow's milk $=105$ litres
Find the H.C.F. of 175 and 105 using Euclid's division method of factorisation method.

$\begin{aligned}
&175=5 \times 5 \times 7 \\
&105=3 \times 5 \times 7
\end{aligned}$
H.C.F. of 175 and $105=5 \times 7=35$
(i) The capacity of the milk can's is 35 litres
(ii) Cows milk $=175$ litres
Number of cans $=\frac{175}{35}=5$

(iii) Buffalow's milk $=105$ litres
Number of cans $=\frac{105}{35}=3$
(i) Capacity of one can $=35$ litres
(ii) Number of can's for cow's milk= 5 litres
(iii) Number of can's for Buffalow's milk $-3$ litres

Question $3 .$
When the positive integers $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ are divided by 13 the respective remainders are 9,7 and 10 . Find the remainder when $a+2 b+3 c$ is divided by 13 .
Answer:
Let the positive integers be $a, b$, and $c$.
$\begin{aligned}
&\mathrm{a}=13 \mathrm{q}+9 \\
&\mathrm{~b}=13 \mathrm{q}+7 \\
&\mathrm{c}=13 \mathrm{q}+10 \\
&\mathrm{a}+2 \mathrm{~b}+3 \mathrm{c}=13 \mathrm{q}+9+2(13 \mathrm{q}+7)+3(13 \mathrm{q}+10) \\
&=13 \mathrm{q}+9+269+14+39 \mathrm{q}+30 \\
&=78 \mathrm{q}+53=(13 \times 6) \mathrm{q}+53
\end{aligned}$
The remainder is 53 .
But $53=13 \times 4+1$
$\therefore$ The remainder is 1

Question $4 .$
Show that 107 is of the form $4 \mathrm{q}+3$ for any integer $\mathrm{q}$.
Answer:

$107=4 \times 26+3$
This is in the form of $a=b q+r$
Hence it is proved.

Question $5 .$
If $(m+1)^{\text {th }}$ term of an A.P. is twice the $(n+1)^{\text {th }}$ term, then prove that $(3 m+1)^{\text {th }}$ term is twice the $(m+n+1)^{\text {th }}$ term.
Solution:
$\begin{aligned}
&t_{n}=a+(n-1) d \\
&t_{m+1}=a+(m+1-1) d \\
&=a+m d \\
&t_{n+1}=a+(n+1-1) d \\
&=a+n d
\end{aligned}$
$\begin{aligned}
&2\left(t_{n+1}\right)=2(a+n d) \\
&t_{m+1}-2 t_{n+1} \ldots \ldots \ldots \ldots \ldots(1) \\
&\Rightarrow a+m d=2(a+n d) \\
&2 a+2 n d-a-m d=0 \\
&a+(2 n-m) d=0 \\
&t_{(3 m+1)}=a+(3 m+1-1) d \\
&=a+3 m d
\end{aligned}$
$\mathrm{t}_{(\mathrm{m}+\mathrm{n}+1)}=\mathrm{a}+(\mathrm{m}+\mathrm{n}+1-1) \mathrm{d}$ $-\mathrm{a}+(\mathrm{m}+\mathrm{n}) \mathrm{d}$
$-a+(m+n) d$ $2\left(t_{(m+n+1)}\right)=2(a+(m+n) d)$
$=2 \mathrm{a}+2 \mathrm{md}+2 \mathrm{nd}$
$\begin{aligned}
&t_{(3 m+1)}=2 t_{(m+n+1)} \ldots \ldots \ldots \ldots . \\
&a+3 m d-2 a+2 m d+2 n d \\
&2 a+2 m d+2 n d-a-3 m d=0 \\
&a-m d+2 n d=0
\end{aligned}$

$\mathrm{a}+(2 \mathrm{n}-\mathrm{m}) \mathrm{d}=0$
$\therefore$ It is proved that $\mathrm{t}_{(3 \mathrm{~m}+1)}=2 \mathrm{t}_{(\mathrm{m}+\mathrm{n}+1)}$
 

Question $6 .$
Find the $12^{\text {th }}$ term from the last term of the A.P $-2,-4,-6, \ldots-100$.
Solution:
$n=\frac{l-a}{d}+1=\frac{-100-(-2)}{-2}+1$ $=\frac{-100+2}{-2}+1=\frac{-98}{-2}+1$ $$ n=49+1=50 $$ $12^{\text {th }}$ term from the last $=39^{\text {th }}$ term from the beginning $\therefore \mathrm{t}_{39}=\mathrm{a}+38 \mathrm{~d}$ $=-2+38(-2)$ $=-2-76$ $=-78$
 

Question 7.
Two A.P.'s have the same common difference. The first term of one A.P. is 2 and that of the other is 7. Show that the difference between their $10^{\text {th }}$ terms is the same as the difference between their 21 st terms, which is the same as the difference between any two corresponding terms.
Answer:
Let the common difference for the 2 A.P be "d"
For the first A.P
$a=2, d=d, n=10$
$\mathrm{t}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}$
$t_{10}=2+9 d \ldots$...(1)
For the $2^{\text {nd }}$ A.P
$a=7, d=d n=10$
$\mathrm{t}_{10}=7+(9) \mathrm{d}$
$=7+9 \mathrm{~d} \ldots . .(2)$
Difference between their 10 th term $\Rightarrow(1)-(2)$
$\begin{aligned}
&=2+9 \mathrm{~d}-(7+9 \mathrm{~d}) \\
&=2+9 \mathrm{~d}-7-9 \mathrm{~d}
\end{aligned}$

$=-5$
For first A.P when $\mathrm{n}=21, \mathrm{a}=2, \mathrm{~d}=\mathrm{d}$
$t_{21}=2+20 \mathrm{~d} \ldots \ldots .(3)$
For second A.P when $\mathrm{n}=21, \mathrm{a}=7, \mathrm{~d}=\mathrm{d}$
$\mathrm{t}_{21}=7+20 \mathrm{~d}$
(4)
Difference between the $21^{\text {st }}$ term $\Rightarrow(3)-(4)$
$=2+20 \mathrm{~d}-(7+20 \mathrm{~d})$
$=2+20 \mathrm{~d}-7-20 \mathrm{~d}$
$=-5$
Difference between their 10 th term and $21^{\text {st }}$ term $=-5$
Hence it is proved.
 

Question $8 .$
A man saved $\square 16500$ in ten years. In each year after the first he saved $\square 100$ more than he did in the preceding year. How much did he save in the first year?
Solution:
$\mathrm{S}_{10}=\square 16500$
$\mathrm{a}, \mathrm{a}+\mathrm{d}, \mathrm{a}+2 \mathrm{~d} \ldots$
$\mathrm{d}=100$
$\mathrm{n}=10$
$\mathrm{S}_{\mathrm{n}}=\frac{n}{2}(2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d})$
$\mathrm{S}_{10}=16500$

$\begin{aligned}
&S_{10}=\frac{10}{2}(2 \times a+9 \times 100) \\
&16500=5(9 a+900) \\
&16500=10 a+4500 \\
&10 a=16500-4500 \\
&10 a=12000 \\
&a=\frac{12000}{10}=\square 1200
\end{aligned}$
$\therefore$ He saved $\square 1200$ in the first year
 

Question $9 .$
Find the G.P. in which the 2 nd term is $\sqrt{6}$ and the 6 th term is $9 \sqrt{6}$.
Solution:
$\begin{aligned}
t_{2} &=\sqrt{6} \\
t_{6} &=9 \sqrt{6} \\
t_{n} &=a r^{n-1} \text { in G.P } \\
\therefore t_{2} &=a r^{2-1}=\sqrt{6} \\
a r &=\sqrt{6}
\end{aligned}$

$\begin{aligned}
a r &=\sqrt{6} \\
t_{6} &=a r^{6-1}=9 \sqrt{6} \\
a r^{5} &=9 \sqrt{6} \\
\frac{(2)}{(1)} &=\frac{a r^{s^{4}}}{a r}=\frac{9 \sqrt{6}}{\sqrt{6}} \\
r^{4}=9 \Rightarrow r^{2} &=3 \Rightarrow r=\sqrt{3}
\end{aligned}$
Substitute $r=\sqrt{3}$ in (1)
$\begin{aligned}
a r &=\sqrt{6} \\
a \sqrt{3} &=\sqrt{6} \\
a &=\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{\frac{6}{3}}=\sqrt{2} \\
\therefore \mathrm{G} P=a, a r, a r^{2} & \ldots \\
&=\sqrt{2}, \sqrt{6}, \sqrt{2} \sqrt{3}^{2}, \cdots \\
&=\sqrt{2}, \sqrt{6}, 3 \sqrt{2}, \ldots
\end{aligned}$

Question $10 .$
The value of a motorcycle depreciates at the rate of $15 \%$ per year. What will be the value of the motor cycle 3 year hence, which is now purchased for $\square 45,000$ ?
Answer:
Value of the motor cycyle $=\square 45000$
a=45000

$Depreciation $=15 \%$ of the cost value
$\begin{aligned}
&=\frac{15}{100} \times 45000 \\
&=15 \times 450
\end{aligned}$

$=6750$
$\mathrm{d}=-6750$ (decrease it is depreciation
Value of the motor cycle lightning of the 2 nd year $=45000-6750$ $=\square 38250$
Depreciation for the 2 nd year $=\frac{15}{100} \times 38250$ $=\square 57370.50$