Samacheer Kalvi 10th Maths Unit Exercise 2 Solutions

10th Maths Unit Exercise 2 solutions (Chapter 2 Numbers and Sequences) with step-by-step answers for all 12 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance8 answers · full steps below

Q2 (i)Capacity of a can $=35$ litres
Q2 (ii) $5$ cans
Q2 (iii) $3$ cans
Q3Remainder $=1$
Q6 $-78$
Q8₹ $1200$ in the first year
Q9 $\sqrt2,\ \sqrt6,\ 3\sqrt2,\ 3\sqrt6,\ldots$ (or) $-\sqrt2,\ \sqrt6,\ -3\sqrt2,\ 3\sqrt6,\ldots$
Q10₹ $27635.63$ (approximately)

Q1. Prove that $n^2 - n$ divisible by $2$ for every positive integer $n$ .

Show steps (7)

Every positive integer $n$ is of the form $2q$ or $2q+1$ (Euclid's division lemma with divisor $2$ ).
$n^2-n=n(n-1)$ .
Case 1: $n=2q$ .
$n^2-n=2q(2q-1)=2\big(q(2q-1)\big)$ , a multiple of $2$ .
Case 2: $n=2q+1$ .
$n^2-n=(2q+1)(2q)=2\big(q(2q+1)\big)$ , a multiple of $2$ .
In both cases $n^2-n$ is divisible by $2$ .

Answer

$n^2-n$ is divisible by $2$ for every positive integer $n$ .

Q2. A milk man has $175$ litres of cow's milk and $105$ litres …

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.

Show steps (6)

The can capacity must divide both $175$ and $105$ , and it must be the largest such number.
So capacity $=\text{HCF}(175,105)$ .
$175=105\times 1+70$
$105=70\times 1+35$
$70=35\times 2+0$
The last non‑zero remainder is $35$ .

Answer

Capacity of a can $=35$ litres

(ii)

Number of cans of cow's milk.

Number of cans $=\dfrac{\text{quantity of cow's milk}}{\text{capacity}}$ .
$=\dfrac{175}{35}=5$

Answer

$5$ cans

(iii)

Number of cans of buffalow's milk.

Number of cans $=\dfrac{105}{35}=3$

Answer

$3$ cans

Q3. When the positive integers $a$ , $b$ and $c$ are divided by …

When the positive integers $a$ , $b$ and $c$ are divided by $13$ the respective remainders are $9$ , $7$ and $10$ . Find the remainder when $a + 2b + 3c$ is divided by $13$ .

Show steps (7)

By Euclid's division lemma,
$a=13q_1+9$ , $b=13q_2+7$ , $c=13q_3+10$ .
$a+2b+3c=13q_1+9+2(13q_2+7)+3(13q_3+10)$
$=13(q_1+2q_2+3q_3)+9+14+30$
$=13(q_1+2q_2+3q_3)+53$
Now $53=13\times 4+1$ .
$a+2b+3c=13(q_1+2q_2+3q_3+4)+1$

Answer

Remainder $=1$

Q4. Show that $107$ is of the form $4q + 3$ for any integer $q$ …

Show that $107$ is of the form $4q + 3$ for any integer $q$ .

Divide $107$ by $4$ using Euclid's division lemma.
$107=4\times 26+3$
This is of the form $4q+3$ with $q=26$ .

Answer

$107=4(26)+3$ , so $107$ is of the form $4q+3$ where $q=26$ .

Q5. If $(m + 1)^{th}$ term of an A.P. is twice the $(n + 1)^{th}$ term, then …

If $(m + 1)^{th}$ term of an A.P. is twice the $(n + 1)^{th}$ term, then prove that $(3m+1)^{th}$ term is twice the $(m + n + 1)^{th}$ term.

Show steps (8)

Let the A.P. have first term $a$ and common difference $d$ . Then $t_k=a+(k-1)d$ .
Given: $t_{m+1}=2\,t_{n+1}$
$a+md=2(a+nd)$
$a+md=2a+2nd$
$a=(m-2n)d$ … (1)
Now $t_{3m+1}=a+3md=(m-2n)d+3md=(4m-2n)d$
And $t_{m+n+1}=a+(m+n)d=(m-2n)d+(m+n)d=(2m-n)d$
$2\,t_{m+n+1}=2(2m-n)d=(4m-2n)d=t_{3m+1}$

Answer

$t_{3m+1}=2\,t_{m+n+1}$ , hence proved.

Q6. Find the $12^{th}$ term from the last term of the A. P …

Find the $12^{th}$ term from the last term of the A. P $-2, -4, -6, \ldots -100$ .

A.P.: $-2,-4,-6,\ldots,-100$ , last term $l=-100$ , $d=-2$ .
$n^{th}$ term from the last $=l-(n-1)d$ .
$12^{th}$ term from last $=-100-(12-1)(-2)$
$=-100+22$
$=-78$

Answer

$-78$

Q7. Two A.P.'s have the same common difference. The first term of one …

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 $10^{th}$ terms is the same as the difference between their $21^{st}$ terms, which is the same as the difference between any two corresponding terms.

Show steps (8)

Let the common difference of both A.P.s be $d$ .
First A.P.: $a=2$ ; Second A.P.: $A=7$ .
$10^{th}$ terms: $t_{10}=2+9d$ , $T_{10}=7+9d$
Difference $=(7+9d)-(2+9d)=5$
$21^{st}$ terms: $t_{21}=2+20d$ , $T_{21}=7+20d$
Difference $=(7+20d)-(2+20d)=5$
Any $n^{th}$ terms: $\big(7+(n-1)d\big)-\big(2+(n-1)d\big)=5$
All the differences equal $5$ , the difference of the first terms.

Answer

The difference is $5$ in every case, so the difference between any two corresponding terms is the same.

Q8. A man saved ₹ $16500$ in ten years. In each year after the …

A man saved ₹ $16500$ in ten years. In each year after the first he saved ₹ $100$ more than he did in the preceding year. How much did he save in the first year?

Show steps (7)

Savings form an A.P. with $d=100$ , $n=10$ , $S_{10}=16500$ .
$S_n=\dfrac{n}{2}\big[2a+(n-1)d\big]$
$16500=\dfrac{10}{2}\big[2a+9(100)\big]$
$16500=5(2a+900)$
$3300=2a+900$
$2a=2400$
$a=1200$

Answer

₹ $1200$ in the first year

Q9. Find the G.P. in which the $2^{nd}$ term is $\sqrt{6}$ and the …

Find the G.P. in which the $2^{nd}$ term is $\sqrt{6}$ and the $6^{th}$ term is $9\sqrt{6}$ .

Show steps (8)

Let the G.P. be $a,ar,ar^2,\ldots$
$t_2=ar=\sqrt{6}$ … (1)
$t_6=ar^5=9\sqrt{6}$ … (2)
$\dfrac{(2)}{(1)}:\ r^4=9\Rightarrow r^2=3\Rightarrow r=\pm\sqrt{3}$
If $r=\sqrt{3}$ : $a=\dfrac{\sqrt6}{\sqrt3}=\sqrt2$
G.P.: $\sqrt2,\ \sqrt6,\ 3\sqrt2,\ 3\sqrt6,\ldots$
If $r=-\sqrt{3}$ : $a=\dfrac{\sqrt6}{-\sqrt3}=-\sqrt2$
G.P.: $-\sqrt2,\ \sqrt6,\ -3\sqrt2,\ 3\sqrt6,\ldots$

Answer

$\sqrt2,\ \sqrt6,\ 3\sqrt2,\ 3\sqrt6,\ldots$ (or) $-\sqrt2,\ \sqrt6,\ -3\sqrt2,\ 3\sqrt6,\ldots$

Q10. The value of a motor cycle depreciates at the rate of $15\%$ …

The value of a motor cycle 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 ₹ $45,000$ ?

Depreciation of $15\%$ per year means each year the value becomes $85\%$ of the previous value.
The values form a G.P. with $a=45000$ and $r=\dfrac{85}{100}=0.85$ .
Value after $3$ years $=45000\times(0.85)^3$
$=45000\times 0.614125$
$=27635.625$

Answer

₹ $27635.63$ (approximately)

Quick check

The least number that is divisible by all the numbers from $1$ to $10$ (both inclusive) is
1. (1) $2025$
2. (2) $5220$
3. (3) $5025$
4. (4) $2520$
Show answer
Answer: (4) $2520$
1. The least such number is the LCM of $1,2,\ldots,10$ .
2. $\text{LCM} = 2^3 \times 3^2 \times 5 \times 7 = 2520$ .
Question 1 of 5
The next term of the sequence $\dfrac{3}{16}, \dfrac{1}{8}, \dfrac{1}{12}, \dfrac{1}{18}, \cdots$ is
1. (1) $\dfrac{1}{24}$
2. (2) $\dfrac{1}{27}$
3. (3) $\dfrac{2}{3}$
4. (4) $\dfrac{1}{81}$
Show answer
Answer: (2) $\dfrac{1}{27}$
1. $\dfrac{1/8}{3/16} = \dfrac{16}{24} = \dfrac{2}{3}$ , $\dfrac{1/12}{1/8} = \dfrac{2}{3}$ , $\dfrac{1/18}{1/12} = \dfrac{2}{3}$ .
2. It is a G.P. with $r = \dfrac{2}{3}$ .
3. Next term $= \dfrac{1}{18} \times \dfrac{2}{3} = \dfrac{1}{27}$ .
Question 2 of 5
The first term of an arithmetic progression is unity and the common difference is $4$ . Which of the following will be a term of this A.P.
1. (1) $4551$
2. (2) $10091$
3. (3) $7881$
4. (4) $13531$
Show answer
Answer: (3) $7881$
1. $a = 1$ , $d = 4$ , so $t_n = 1 + (n-1)4 = 4n - 3$ .
2. Every term leaves remainder $1$ when divided by $4$ .
3. $7881 = 4 \times 1971 - 3$ , so $7881$ is the $1971^{\text{st}}$ term.
Question 3 of 5
If the HCF of $65$ and $117$ is expressible in the form of $65m - 117$ , then the value of $m$ is
1. (1) $4$
2. (2) $2$
3. (3) $1$
4. (4) $3$
Show answer
Answer: (2) $2$
1. $117 = 65 \times 1 + 52$ , $65 = 52 \times 1 + 13$ , $52 = 13 \times 4 + 0$ .
2. So HCF $(65,117) = 13$ .
3. $65m - 117 = 13 \Rightarrow 65m = 130 \Rightarrow m = 2$ .
Question 4 of 5
In an A.P., the first term is $1$ and the common difference is $4$ . How many terms of the A.P. must be taken for their sum to be equal to $120$ ?
1. (1) $6$
2. (2) $7$
3. (3) $8$
4. (4) $9$
Show answer
Answer: (3) $8$
1. $S_n = \dfrac{n}{2}[2(1) + (n-1)4] = \dfrac{n}{2}(4n - 2) = n(2n-1)$ .
2. $n(2n-1) = 120 \Rightarrow 2n^2 - n - 120 = 0$ .
3. $(n-8)(2n+15) = 0 \Rightarrow n = 8$ .
Question 5 of 5

Chapter 2 formulas

Euclid's division lemma
$a = bq + r,\ 0 \le r < |b|$
Fundamental theorem of arithmetic
$N = p_1^{a_1} p_2^{a_2} p_3^{a_3} \cdots p_k^{a_k}\ \text{(unique except for order)}$
Arithmetic Progression
$a,\ a+d,\ a+2d,\ a+3d,\ \ldots$
n-th term of an A.P.
$t_n = a + (n-1)d$
Sum to first n terms of an A.P.
$S_n = \frac{n}{2}\left[2a + (n-1)d\right]$
Sum of an A.P. when the last term l is given
$S_n = \frac{n}{2}\left[a + l\right]$
Geometric Progression
$a,\ ar,\ ar^2,\ \ldots,\ ar^{n-1}$
n-th term of a G.P.
$t_n = ar^{n-1}$
Sum to first n terms of a G.P. (r ≠ 1)
$S_n = \frac{a(r^n - 1)}{r - 1}$
Sum to first n terms of a G.P. when r = 1
$S_n = na$
Sum to infinite terms of a G.P. (−1 < r < 1)
$S_\infty = \frac{a}{1 - r}$
Sum of first n natural numbers
$1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$
Sum of squares of first n natural numbers
$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$
Sum of cubes of first n natural numbers
$1^3 + 2^3 + 3^3 + \cdots + n^3 = \left[\frac{n(n+1)}{2}\right]^2$
Sum of first n odd natural numbers
$1 + 3 + 5 + \cdots + (2n-1) = n^2$

Answers checked against the Samacheer Kalvi textbook · Updated 28 September 2026

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Q1. Prove that n2−nn^2 - nn2−n divisible by 222 for every positive integer nnn .

Q2. A milk man has 175175175 litres of cow's milk and 105105105 litres …

(i)

(ii)

(iii)

Q3. When the positive integers aaa , bbb and ccc are divided by …

Q4. Show that 107107107 is of the form 4q+34q + 34q+3 for any integer qqq …

Q5. If (m+1)th(m + 1)^{th}(m+1)th term of an A.P. is twice the (n+1)th(n + 1)^{th}(n+1)th term, then …

Q6. Find the 12th12^{th}12th term from the last term of the A. P …

Q7. Two A.P.'s have the same common difference. The first term of one …

Q8. A man saved ₹165001650016500 in ten years. In each year after the …

Q9. Find the G.P. in which the 2nd2^{nd}2nd term is 6\sqrt{6}6​ and the …

Q10. The value of a motor cycle depreciates at the rate of 15%15\%15% …

Quick check

Q1. Prove that $n^2 - n$ divisible by $2$ for every positive integer $n$ .

Q2. A milk man has $175$ litres of cow's milk and $105$ litres …

Q3. When the positive integers $a$ , $b$ and $c$ are divided by …

Q4. Show that $107$ is of the form $4q + 3$ for any integer $q$ …

Q5. If $(m + 1)^{th}$ term of an A.P. is twice the $(n + 1)^{th}$ term, then …

Q6. Find the $12^{th}$ term from the last term of the A. P …

Q8. A man saved ₹ $16500$ in ten years. In each year after the …

Q9. Find the G.P. in which the $2^{nd}$ term is $\sqrt{6}$ and the …

Q10. The value of a motor cycle depreciates at the rate of $15\%$ …