Samacheer Kalvi 10th Maths Exercise 2.9 Solutions

10th Maths Exercise 2.9 solutions (Chapter 2 Numbers and Sequences) with step-by-step answers for all 14 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance14 answers · full steps below

Q1 (i) $1830$
Q1 (ii) $1584$
Q1 (iii) $3003$
Q1 (iv) $1240$
Q1 (v) $3256$
Q1 (vi) $42075$
Q1 (vii) $1296$
Q2 $105625$
Q3 $210$
Q4 $15$ terms
Q5 $n=9$
Q6 $4615$ sq.cm
Q7 (i) $n^2(4n+3)=4n^3+3n^2$
Q7 (ii) $2240$

Q1. Find the sum of the following series:

(i)

$1 + 2 + 3 + \cdots + 60$

Use $1+2+\cdots+n=\dfrac{n(n+1)}{2}$ with $n=60$ .
$=\dfrac{60\times 61}{2}=1830$

Answer

$1830$

(ii)

$3 + 6 + 9 + \cdots + 96$

Take $3$ common: $3+6+9+\cdots+96=3(1+2+3+\cdots+32)$
$=3\times\dfrac{32\times 33}{2}=3\times 528$
$=1584$

Answer

$1584$

(iii)

$51 + 52 + 53 + \cdots + 92$

$51+52+\cdots+92=(1+2+\cdots+92)-(1+2+\cdots+50)$
$=\dfrac{92\times 93}{2}-\dfrac{50\times 51}{2}$
$=4278-1275=3003$

Answer

$3003$

(iv)

$1 + 4 + 9 + 16 + \cdots + 225$

The terms are $1^2+2^2+3^2+\cdots+15^2$ (since $225=15^2$ ).
Use $\dfrac{n(n+1)(2n+1)}{6}$ with $n=15$ .
$=\dfrac{15\times 16\times 31}{6}=1240$

Answer

$1240$

(v)

$6^2 + 7^2 + 8^2 + \cdots + 21^2$

$6^2+7^2+\cdots+21^2=(1^2+\cdots+21^2)-(1^2+\cdots+5^2)$
$=\dfrac{21\times 22\times 43}{6}-\dfrac{5\times 6\times 11}{6}$
$=3311-55=3256$

Answer

$3256$

(vi)

$10^3 + 11^3 + 12^3 + \cdots + 20^3$

$10^3+11^3+\cdots+20^3=(1^3+\cdots+20^3)-(1^3+\cdots+9^3)$
$=\left(\dfrac{20\times 21}{2}\right)^2-\left(\dfrac{9\times 10}{2}\right)^2$
$=210^2-45^2=44100-2025$
$=42075$

Answer

$42075$

(vii)

$1 + 3 + 5 + \cdots + 71$

This is odd numbers with $a=1,\ d=2,\ l=71$ .
Number of terms $n=\dfrac{l-a}{d}+1=\dfrac{71-1}{2}+1=36$
Sum of first $n$ odd numbers $=n^2=36^2=1296$

Answer

$1296$

Q2. If $1 + 2 + 3 + \cdots + k = 325$ , then find $1^3 + 2^3 + 3^3 + \cdots + k^3$ .

$1^3+2^3+\cdots+k^3=\left(\dfrac{k(k+1)}{2}\right)^2=(1+2+\cdots+k)^2$
$=(325)^2$
$=105625$

Answer

$105625$

Q3. If $1^3 + 2^3 + 3^3 + \cdots + k^3 = 44100$ then find $1 + 2 + 3 + \cdots + k$ .

$(1+2+\cdots+k)^2=1^3+2^3+\cdots+k^3=44100$
$1+2+\cdots+k=\sqrt{44100}=210$

Answer

$210$

Q4. How many terms of the series $1^3 + 2^3 + 3^3 + \cdots$ should be taken to get …

How many terms of the series $1^3 + 2^3 + 3^3 + \cdots$ should be taken to get the sum $14400$ ?

$\left(\dfrac{n(n+1)}{2}\right)^2=14400$
$\dfrac{n(n+1)}{2}=120 \Rightarrow n^2+n-240=0$
$(n+16)(n-15)=0$
$n=15$ (as $n$ is a natural number)

Answer

$15$ terms

Q5. The sum of the cubes of the first $n$ natural numbers is …

The sum of the cubes of the first $n$ natural numbers is $2025$ , then find the value of $n$ .

$\left(\dfrac{n(n+1)}{2}\right)^2=2025$
$\dfrac{n(n+1)}{2}=45 \Rightarrow n^2+n-90=0$
$(n+10)(n-9)=0$
$n=9$ (as $n$ is a natural number)

Answer

$n=9$

Q6. Rekha has $15$ square colour papers of sizes $10$ cm, $11$ cm …

Rekha has $15$ square colour papers of sizes $10$ cm, $11$ cm, $12$ cm,…, $24$ cm. How much area can be decorated with these colour papers?

Total area $=10^2+11^2+\cdots+24^2$
$=(1^2+\cdots+24^2)-(1^2+\cdots+9^2)$
$=\dfrac{24\times 25\times 49}{6}-\dfrac{9\times 10\times 19}{6}$
$=4900-285=4615$

Answer

$4615$ sq.cm

Q7. Find the sum of the series $(2^3 - 1^3) + (4^3 - 3^3) + (6^3 - 5^3) + \cdots$ to

(i)

$n$ terms

Show steps (7)

The $k$ -th term is $(2k)^3-(2k-1)^3$ .
$(2k)^3-(2k-1)^3=8k^3-(8k^3-12k^2+6k-1)=12k^2-6k+1$
Sum $=12\sum k^2-6\sum k+\sum 1$
$=12\times\dfrac{n(n+1)(2n+1)}{6}-6\times\dfrac{n(n+1)}{2}+n$
$=2n(n+1)(2n+1)-3n(n+1)+n$
$=n\left[2(2n^2+3n+1)-3n-3+1\right]$
$=n\left[4n^2+3n\right]=n^2(4n+3)$

Answer

$n^2(4n+3)=4n^3+3n^2$

(ii)

$8$ terms

Put $n=8$ in $n^2(4n+3)$ .
$=8^2(4\times 8+3)=64\times 35$
$=2240$

Answer

$2240$

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. Find the sum of the following series:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Q2. If 1+2+3+⋯+k=3251 + 2 + 3 + \cdots + k = 3251+2+3+⋯+k=325 , then find 13+23+33+⋯+k31^3 + 2^3 + 3^3 + \cdots + k^313+23+33+⋯+k3 .

Q3. If 13+23+33+⋯+k3=441001^3 + 2^3 + 3^3 + \cdots + k^3 = 4410013+23+33+⋯+k3=44100 then find 1+2+3+⋯+k1 + 2 + 3 + \cdots + k1+2+3+⋯+k .

Q4. How many terms of the series 13+23+33+⋯1^3 + 2^3 + 3^3 + \cdots13+23+33+⋯ should be taken to get …

Q5. The sum of the cubes of the first nnn natural numbers is …

Q6. Rekha has 151515 square colour papers of sizes 101010 cm, 111111 cm …

Q7. Find the sum of the series (23−13)+(43−33)+(63−53)+⋯(2^3 - 1^3) + (4^3 - 3^3) + (6^3 - 5^3) + \cdots(23−13)+(43−33)+(63−53)+⋯ to

(i)

(ii)

Quick check

Q2. If $1 + 2 + 3 + \cdots + k = 325$ , then find $1^3 + 2^3 + 3^3 + \cdots + k^3$ .

Q3. If $1^3 + 2^3 + 3^3 + \cdots + k^3 = 44100$ then find $1 + 2 + 3 + \cdots + k$ .

Q4. How many terms of the series $1^3 + 2^3 + 3^3 + \cdots$ should be taken to get …

Q5. The sum of the cubes of the first $n$ natural numbers is …

Q6. Rekha has $15$ square colour papers of sizes $10$ cm, $11$ cm …

Q7. Find the sum of the series $(2^3 - 1^3) + (4^3 - 3^3) + (6^3 - 5^3) + \cdots$ to