10th Maths Chapter 2 Numbers and Sequences

10th Maths Chapter 2 Numbers and Sequences solutions, with the chapter's formulas and every exercise worked step by step, as per the 2026-27 Samacheer Kalvi textbook.

This chapter looks at whole numbers and patterns in lists of numbers. You study division with quotient and remainder, highest common factor, prime factorisation, and simple work with remainders. Then come sequences: arithmetic ones with a fixed difference, geometric ones with a fixed ratio, and their sums. The exercises ask you to find terms, add series, prove small results, and solve word problems on savings, growth and decay.

Exercises in this chapter

Key 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
$1 + 3 + 5 + \cdots + (2n-1) = n^2$

Chapter 2 one-mark practice

Euclid's division lemma states that for positive integers $a$ and $b$ , there exist unique integers $q$ and $r$ such that $a = bq + r$ , where $r$ must satisfy.
1. (1) $1 < r < b$
2. (2) $0 < r < b$
3. (3) $0 \leq r < b$
4. (4) $0 < r \leq b$
Show answer
Answer: (3) $0 \leq r < b$
1. In $a = bq + r$ , the remainder $r$ can be zero (when $b$ divides $a$ ) but must be less than the divisor $b$ .
2. So $0 \le r < b$ .
Question 1 of 10
Using Euclid's division lemma, if the cube of any positive integer is divided by $9$ then the possible remainders are
1. (1) $0, 1, 8$
2. (2) $1, 4, 8$
3. (3) $0, 1, 3$
4. (4) $1, 3, 5$
Show answer
Answer: (1) $0, 1, 8$
1. Any positive integer is of the form $3q$ , $3q+1$ or $3q+2$ .
2. $(3q)^3 = 27q^3 = 9(3q^3) \Rightarrow$ remainder $0$ .
3. $(3q+1)^3 = 27q^3+27q^2+9q+1 = 9(3q^3+3q^2+q)+1 \Rightarrow$ remainder $1$ .
4. $(3q+2)^3 = 27q^3+54q^2+36q+8 = 9(3q^3+6q^2+4q)+8 \Rightarrow$ remainder $8$ .
Question 2 of 10
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 3 of 10
The sum of the exponents of the prime factors in the prime factorization of $1729$ is
1. (1) $1$
2. (2) $2$
3. (3) $3$
4. (4) $4$
Show answer
Answer: (3) $3$
1. $1729 = 7 \times 13 \times 19 = 7^1 \times 13^1 \times 19^1$ .
2. Sum of exponents $= 1 + 1 + 1 = 3$ .
Question 4 of 10
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 5 of 10
$7^{4k} \equiv$ _____ $(\text{mod } 100)$
1. (1) $1$
2. (2) $2$
3. (3) $3$
4. (4) $4$
Show answer
Answer: (1) $1$
1. $7^2 = 49$ , so $7^4 = 49^2 = 2401$ .
2. $2401 = 24 \times 100 + 1 \Rightarrow 7^4 \equiv 1 \pmod{100}$ .
3. Hence $7^{4k} = (7^4)^k \equiv 1^k \equiv 1 \pmod{100}$ .
Question 6 of 10
Given $F_1 = 1$ , $F_2 = 3$ and $F_n = F_{n-1} + F_{n-2}$ then $F_5$ is
1. (1) $3$
2. (2) $5$
3. (3) $8$
4. (4) $11$
Show answer
Answer: (4) $11$
1. $F_3 = F_2 + F_1 = 3 + 1 = 4$ .
2. $F_4 = F_3 + F_2 = 4 + 3 = 7$ .
3. $F_5 = F_4 + F_3 = 7 + 4 = 11$ .
Question 7 of 10
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 8 of 10
If $6$ times of $6^{\text{th}}$ term of an A.P. is equal to $7$ times the $7^{\text{th}}$ term, then the $13^{\text{th}}$ term of the A.P. is
1. (1) $0$
2. (2) $6$
3. (3) $7$
4. (4) $13$
Show answer
Answer: (1) $0$
1. $6t_6 = 7t_7 \Rightarrow 6(a+5d) = 7(a+6d)$ .
2. $6a + 30d = 7a + 42d \Rightarrow a + 12d = 0$ .
3. $t_{13} = a + 12d = 0$ .
Question 9 of 10
An A.P. consists of $31$ terms. If its $16^{\text{th}}$ term is $m$ , then the sum of all the terms of this A.P. is
1. (1) $16\,m$
2. (2) $62\,m$
3. (3) $31\,m$
4. (4) $\dfrac{31}{2}\,m$
Show answer
Answer: (3) $31\,m$
1. $t_{16} = a + 15d = m$ .
2. $S_{31} = \dfrac{31}{2}\,[2a + 30d] = \dfrac{31}{2} \times 2(a + 15d)$ .
3. $= 31(a+15d) = 31m$ .
Question 10 of 10

Spotted a mistake? Email us at [email protected] and we will fix it.