Samacheer Kalvi 10th Maths Exercise 2.5 Solutions

10th Maths Exercise 2.5 solutions (Chapter 2 Numbers and Sequences) with step-by-step answers for all 21 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance20 answers · full steps below

Q1 (i)Yes, it is an A.P. with $d=-2$
Q1 (ii)No, it is not an A.P.
Q1 (iii)Yes, it is an A.P. with $d=4$
Q1 (iv)Yes, it is an A.P. with $d=\dfrac{1}{3}$
Q1 (v)No, it is not an A.P.
Q2 (i) $5, 11, 17, 23, \ldots$
Q2 (ii) $7, 2, -3, -8, \ldots$
Q2 (iii) $\dfrac{3}{4}, \dfrac{5}{4}, \dfrac{7}{4}, \dfrac{9}{4}, \ldots$
Q3 (i) $a=-1$ , $d=2$
Q3 (ii) $a=-3$ , $d=-7$
Q4 $t_{19}=-83$
Q5 $-54$ is the $15^{th}$ term
Q6The middle terms are $93$ and $99$
Q8 $k=4$
Q9 $x=3$ , $y=17$ , $z=31$
Q10 $78$ seats in the last row
Q11The three terms are $2, 9, 16$
Q12 $t_9:t_{13}=5:7$
Q13Temperatures are $-3^\circ$ C, $0^\circ$ C, $3^\circ$ C, $6^\circ$ C, $9^\circ$ C (Monday to Friday)
Q14She saves ₹20,000 per month in the $31^{st}$ year, i.e. after $30$ years

Q1. Check whether the following sequences are in A.P.

(i)

$a - 3,\ a - 5,\ a - 7,...$

$t_2-t_1=(a-5)-(a-3)=-2$
$t_3-t_2=(a-7)-(a-5)=-2$
The differences are equal.

Answer

Yes, it is an A.P. with $d=-2$

(ii)

$\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}, \dfrac{1}{5},...$

$t_2-t_1=\dfrac{1}{3}-\dfrac{1}{2}=-\dfrac{1}{6}$
$t_3-t_2=\dfrac{1}{4}-\dfrac{1}{3}=-\dfrac{1}{12}$
$t_2-t_1\neq t_3-t_2$

Answer

No, it is not an A.P.

(iii)

$9,\ 13,\ 17,\ 21,\ 25,...$

$t_2-t_1=13-9=4$
$t_3-t_2=17-13=4$
$t_4-t_3=21-17=4$ , $t_5-t_4=25-21=4$
All differences are equal.

Answer

Yes, it is an A.P. with $d=4$

(iv)

$\dfrac{-1}{3}, 0, \dfrac{1}{3}, \dfrac{2}{3},...$

$t_2-t_1=0-\left(-\dfrac{1}{3}\right)=\dfrac{1}{3}$
$t_3-t_2=\dfrac{1}{3}-0=\dfrac{1}{3}$
$t_4-t_3=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}$
All differences are equal.

Answer

Yes, it is an A.P. with $d=\dfrac{1}{3}$

(v)

$1, -1, 1, -1, 1, -1,...$

$t_2-t_1=-1-1=-2$
$t_3-t_2=1-(-1)=2$
$t_2-t_1\neq t_3-t_2$

Answer

No, it is not an A.P.

Q2. First term $a$ and common difference $d$ are given below. Find the …

First term $a$ and common difference $d$ are given below. Find the corresponding A.P.

(i)

$a = 5$ , $d = 6$

A.P. is $a,\ a+d,\ a+2d,\ a+3d,\ldots$
$=5,\ 5+6,\ 5+12,\ 5+18,\ldots$

Answer

$5, 11, 17, 23, \ldots$

(ii)

$a = 7$ , $d = -5$

A.P. is $a,\ a+d,\ a+2d,\ a+3d,\ldots$
$=7,\ 7-5,\ 7-10,\ 7-15,\ldots$

Answer

$7, 2, -3, -8, \ldots$

(iii)

$a = \dfrac{3}{4}$ , $d = \dfrac{1}{2}$

A.P. is $a,\ a+d,\ a+2d,\ a+3d,\ldots$
$=\dfrac{3}{4},\ \dfrac{3}{4}+\dfrac{1}{2},\ \dfrac{3}{4}+1,\ \dfrac{3}{4}+\dfrac{3}{2},\ldots$

Answer

$\dfrac{3}{4}, \dfrac{5}{4}, \dfrac{7}{4}, \dfrac{9}{4}, \ldots$

Q3. Find the first term and common difference of the Arithmetic Progressions whose …

Find the first term and common difference of the Arithmetic Progressions whose $n^{th}$ terms are given below

(i)

$t_n = -3 + 2n$

$t_n=-3+2n$
$t_1=-3+2=-1$
$t_2=-3+4=1$
$d=t_2-t_1=1-(-1)=2$

Answer

$a=-1$ , $d=2$

(ii)

$t_n = 4 - 7n$

$t_n=4-7n$
$t_1=4-7=-3$
$t_2=4-14=-10$
$d=t_2-t_1=-10-(-3)=-7$

Answer

$a=-3$ , $d=-7$

Q4. Find the $19^{th}$ term of an A.P. $-11, -15, -19,...$

$a=-11$ , $d=-15-(-11)=-4$
$t_n=a+(n-1)d$
$t_{19}=-11+18(-4)$
$=-11-72=-83$

Answer

$t_{19}=-83$

Q5. Which term of an A.P. $16, 11, 6, 1,...$ is $-54$ ?

$a=16$ , $d=11-16=-5$
$t_n=a+(n-1)d=16+(n-1)(-5)=21-5n$
$21-5n=-54$
$5n=75\Rightarrow n=15$

Answer

$-54$ is the $15^{th}$ term

Q6. Find the middle term(s) of an A.P. $9, 15, 21, 27,\ldots,183$ .

$a=9$ , $d=15-9=6$ , $l=183$
$n=\dfrac{l-a}{d}+1=\dfrac{183-9}{6}+1=30$
$n$ is even, so there are two middle terms: the $15^{th}$ and $16^{th}$
$t_{15}=9+14(6)=93$
$t_{16}=9+15(6)=99$

Answer

The middle terms are $93$ and $99$

Q7. If nine times ninth term is equal to the fifteen times fifteenth …

If nine times ninth term is equal to the fifteen times fifteenth term, show that six times twenty fourth term is zero.

Show steps (7)

Given $9t_9=15t_{15}$
$9(a+8d)=15(a+14d)$
$9a+72d=15a+210d$
$0=6a+138d$
$6(a+23d)=0$
But $a+23d=t_{24}$
So $6t_{24}=0$

Answer

$6t_{24}=0$ , hence six times the $24^{th}$ term is zero

Q8. If $3 + k,\ 18 - k,\ 5k + 1$ are in A.P. then find $k$ .

Three numbers are in A.P. if $2b=a+c$
$2(18-k)=(3+k)+(5k+1)$
$36-2k=6k+4$
$32=8k$
$k=4$

Answer

$k=4$

Q9. Find $x$ , $y$ and $z$ , given that the numbers $x, 10, y, 24, z$ …

Find $x$ , $y$ and $z$ , given that the numbers $x, 10, y, 24, z$ are in A.P.

Show steps (6)

$10$ is $t_2$ and $24$ is $t_4$
$t_4-t_2=2d\Rightarrow 24-10=2d$
$d=7$
$x=10-7=3$
$y=10+7=17$
$z=24+7=31$

Answer

$x=3$ , $y=17$ , $z=31$

Q10. In a theatre, there are $20$ seats in the front row and …

In a theatre, there are $20$ seats in the front row and $30$ rows were allotted. Each successive row contains two additional seats than its front row. How many seats are there in the last row?

Seats form an A.P. with $a=20$ , $d=2$ , $n=30$
$t_{30}=a+29d$
$=20+29(2)=20+58=78$

Answer

$78$ seats in the last row

Q11. The sum of three consecutive terms that are in A.P. is $27$ …

The sum of three consecutive terms that are in A.P. is $27$ and their product is $288$ . Find the three terms.

Show steps (6)

Take the terms as $a-d,\ a,\ a+d$
Sum: $3a=27\Rightarrow a=9$
Product: $(9-d)(9)(9+d)=288$
$81-d^2=32$
$d^2=49\Rightarrow d=\pm 7$
If $d=7$ : $2, 9, 16$ ; if $d=-7$ : $16, 9, 2$

Answer

The three terms are $2, 9, 16$

Q12. The ratio of $6^{th}$ and $8^{th}$ term of an A.P. is $7:9$ …

The ratio of $6^{th}$ and $8^{th}$ term of an A.P. is $7:9$ . Find the ratio of $9^{th}$ term to $13^{th}$ term.

$\dfrac{t_6}{t_8}=\dfrac{a+5d}{a+7d}=\dfrac{7}{9}$
$9a+45d=7a+49d$
$2a=4d\Rightarrow a=2d$
$\dfrac{t_9}{t_{13}}=\dfrac{a+8d}{a+12d}=\dfrac{2d+8d}{2d+12d}=\dfrac{10d}{14d}=\dfrac{5}{7}$

Answer

$t_9:t_{13}=5:7$

Q13. In a winter season let us take the temperature of Ooty from …

In a winter season let us take the temperature of Ooty from Monday to Friday to be in A.P. The sum of temperatures from Monday to Wednesday is $0^\circ$ C and the sum of the temperatures from Wednesday to Friday is $18^\circ$ C. Find the temperature on each of the five days.

Let the five temperatures be $a,\ a+d,\ a+2d,\ a+3d,\ a+4d$
Mon + Tue + Wed $=3a+3d=0\Rightarrow a+d=0$ ...(1)
Wed + Thu + Fri $=3a+9d=18\Rightarrow a+3d=6$ ...(2)
(2) $-$ (1): $2d=6\Rightarrow d=3$
From (1): $a=-3$

Answer

Temperatures are $-3^\circ$ C, $0^\circ$ C, $3^\circ$ C, $6^\circ$ C, $9^\circ$ C (Monday to Friday)

Q14. Priya earned ₹ $15{,}000$ in the first month. Thereafter her salary increased by …

Priya earned ₹ $15{,}000$ in the first month. Thereafter her salary increased by ₹ $1500$ per year. Her expenses are ₹ $13{,}000$ during the first month and the expenses increases by ₹ $900$ per year. How long will it take for her to save ₹ $20{,}000$ per month.

Show steps (6)

Saving in the first month $=15000-13000=₹2000$
Yearly increase in saving $=1500-900=₹600$
Savings form an A.P. with $a=2000$ , $d=600$
$t_n=2000+(n-1)600=20000$
$(n-1)600=18000$
$n-1=30\Rightarrow n=31$

Answer

She saves ₹20,000 per month in the $31^{st}$ year, i.e. after $30$ years

Quick check

If $A = 2^{65}$ and $B = 2^{64} + 2^{63} + 2^{62} + \cdots + 2^{0}$ which of the following is true?
1. (1) $B$ is $2^{64}$ more than $A$
2. (2) $A$ and $B$ are equal
3. (3) $B$ is larger than $A$ by $1$
4. (4) $A$ is larger than $B$ by $1$
Show answer
Answer: (4) $A$ is larger than $B$ by $1$
1. $B = 2^0 + 2^1 + \cdots + 2^{64}$ is a G.P. with $a = 1$ , $r = 2$ , $n = 65$ .
2. $B = \dfrac{1(2^{65} - 1)}{2 - 1} = 2^{65} - 1$ .
3. $A - B = 2^{65} - (2^{65} - 1) = 1$ .
Question 1 of 5
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 2 of 5
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 3 of 5
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 4 of 5
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 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. Check whether the following sequences are in A.P.

(i)

(ii)

(iii)

(iv)

(v)

Q2. First term aaa and common difference ddd are given below. Find the …

(i)

(ii)

(iii)

Q3. Find the first term and common difference of the Arithmetic Progressions whose …

(i)

(ii)

Q4. Find the 19th19^{th}19th term of an A.P. −11,−15,−19,...-11, -15, -19,...−11,−15,−19,...

Q5. Which term of an A.P. 16,11,6,1,...16, 11, 6, 1,...16,11,6,1,... is −54-54−54 ?

Q6. Find the middle term(s) of an A.P. 9,15,21,27,…,1839, 15, 21, 27,\ldots,1839,15,21,27,…,183 .

Q7. If nine times ninth term is equal to the fifteen times fifteenth …

Q8. If 3+k, 18−k, 5k+13 + k,\ 18 - k,\ 5k + 13+k, 18−k, 5k+1 are in A.P. then find kkk .

Q9. Find xxx , yyy and zzz , given that the numbers x,10,y,24,zx, 10, y, 24, zx,10,y,24,z …

Q10. In a theatre, there are 202020 seats in the front row and …

Q11. The sum of three consecutive terms that are in A.P. is 272727 …

Q12. The ratio of 6th6^{th}6th and 8th8^{th}8th term of an A.P. is 7:97:97:9 …

Q13. In a winter season let us take the temperature of Ooty from …

Q14. Priya earned ₹15,00015{,}00015,000 in the first month. Thereafter her salary increased by …

Quick check

Q2. First term $a$ and common difference $d$ are given below. Find the …

Q4. Find the $19^{th}$ term of an A.P. $-11, -15, -19,...$

Q5. Which term of an A.P. $16, 11, 6, 1,...$ is $-54$ ?

Q6. Find the middle term(s) of an A.P. $9, 15, 21, 27,\ldots,183$ .

Q8. If $3 + k,\ 18 - k,\ 5k + 1$ are in A.P. then find $k$ .

Q9. Find $x$ , $y$ and $z$ , given that the numbers $x, 10, y, 24, z$ …

Q10. In a theatre, there are $20$ seats in the front row and …

Q11. The sum of three consecutive terms that are in A.P. is $27$ …

Q12. The ratio of $6^{th}$ and $8^{th}$ term of an A.P. is $7:9$ …

Q14. Priya earned ₹ $15{,}000$ in the first month. Thereafter her salary increased by …