Search

Samacheer Kalvi 10th Maths Exercise 2.4 Solutions

10th Maths Exercise 2.4 solutions (Chapter 2 Numbers and Sequences) with step-by-step answers for all 13 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance13 answers · full steps below
  1. Q1 (i)216, 648, 1944216,\ 648,\ 1944
  2. Q1 (ii)−7, −11, −15-7,\ -11,\ -15
  3. Q1 (iii)425, 536, 649\dfrac{4}{25},\ \dfrac{5}{36},\ \dfrac{6}{49}
  4. Q2 (i)−1, 6, 25, 62-1,\ 6,\ 25,\ 62
  5. Q2 (ii)2, −6, 12, −202,\ -6,\ 12,\ -20
  6. Q2 (iii)−4, 2, 12, 26-4,\ 2,\ 12,\ 26
  7. Q3 (i)an=n2+1a_n=n^2+1
  8. Q3 (ii)an=n−1na_n=\dfrac{n-1}{n}
  9. Q3 (iii)an=5n−2a_n=5n-2
  10. Q4 (i)a6=154,a13=133a_6=\dfrac{15}{4},\quad a_{13}=\dfrac{13}{3}
  11. Q4 (ii)a4=−12,a11=−117a_4=-12,\quad a_{11}=-117
  12. Q5a8=6311,a15=22531a_8=\dfrac{63}{11},\quad a_{15}=\dfrac{225}{31}
  13. Q61, 1, 3, 7, 17, 411,\ 1,\ 3,\ 7,\ 17,\ 41

Q1. Find the next three terms of the following sequence.

(i)

8, 24, 72, …8,\ 24,\ 72,\ \ldots

  1. Compare successive terms: 24÷8=324\div 8=3, 72÷24=372\div 24=3.

  2. So each term is multiplied by 33.

  3. a4=72×3=216a_4=72\times 3=216

  4. a5=216×3=648a_5=216\times 3=648

  5. a6=648×3=1944a_6=648\times 3=1944

Answer

216, 648, 1944216,\ 648,\ 1944

(ii)

5, 1, −3, …5,\ 1,\ -3,\ \ldots

  1. 1−5=−41-5=-4, −3−1=−4-3-1=-4.

  2. Each term is decreased by 44.

  3. a4=−3−4=−7a_4=-3-4=-7

  4. a5=−7−4=−11a_5=-7-4=-11

  5. a6=−11−4=−15a_6=-11-4=-15

Answer

−7, −11, −15-7,\ -11,\ -15

(iii)

14, 29, 316, …\dfrac{1}{4},\ \dfrac{2}{9},\ \dfrac{3}{16},\ \ldots

  1. Numerators: 1,2,3,…1,2,3,\ldots so the nthn^{th} numerator is nn.

  2. Denominators: 4,9,16,…4,9,16,\ldots that is 22,32,42,…2^2,3^2,4^2,\ldots so it is (n+1)2(n+1)^2.

  3. Hence an=n(n+1)2a_n=\dfrac{n}{(n+1)^2}.

  4. a4=425a_4=\dfrac{4}{25}, a5=536a_5=\dfrac{5}{36}, a6=649a_6=\dfrac{6}{49}

Answer

425, 536, 649\dfrac{4}{25},\ \dfrac{5}{36},\ \dfrac{6}{49}

Q2. Find the first four terms of the sequences whose nthn^{th} terms are …

Find the first four terms of the sequences whose nthn^{th} terms are given by

(i)

an=n3−2a_n = n^3 - 2

  1. a1=13−2=−1a_1=1^3-2=-1

  2. a2=23−2=6a_2=2^3-2=6

  3. a3=33−2=25a_3=3^3-2=25

  4. a4=43−2=62a_4=4^3-2=62

Answer

−1, 6, 25, 62-1,\ 6,\ 25,\ 62

(ii)

an=(−1)n+1n(n+1)a_n = (-1)^{n+1} n(n+1)

  1. a1=(−1)2 (1)(2)=2a_1=(-1)^{2}\,(1)(2)=2

  2. a2=(−1)3 (2)(3)=−6a_2=(-1)^{3}\,(2)(3)=-6

  3. a3=(−1)4 (3)(4)=12a_3=(-1)^{4}\,(3)(4)=12

  4. a4=(−1)5 (4)(5)=−20a_4=(-1)^{5}\,(4)(5)=-20

Answer

2, −6, 12, −202,\ -6,\ 12,\ -20

(iii)

an=2n2−6a_n = 2n^2 - 6

  1. a1=2(1)2−6=−4a_1=2(1)^2-6=-4

  2. a2=2(2)2−6=2a_2=2(2)^2-6=2

  3. a3=2(3)2−6=12a_3=2(3)^2-6=12

  4. a4=2(4)2−6=26a_4=2(4)^2-6=26

Answer

−4, 2, 12, 26-4,\ 2,\ 12,\ 26

Q3. Find the nthn^{th} term of the following sequences

(i)

2,5,10,17,…2, 5, 10, 17, \ldots

  1. Write each term as one more than a square:

  2. 2=12+12=1^2+1, 5=22+15=2^2+1, 10=32+110=3^2+1, 17=42+117=4^2+1.

  3. So the nthn^{th} term is n2+1n^2+1.

Answer

an=n2+1a_n=n^2+1

(ii)

0,12,23,…0, \dfrac{1}{2}, \dfrac{2}{3}, \ldots

  1. Write the terms as fractions: 01,12,23,…\dfrac{0}{1},\dfrac{1}{2},\dfrac{2}{3},\ldots

  2. Numerator is one less than the denominator, and the denominator is nn.

  3. So an=n−1na_n=\dfrac{n-1}{n}.

Answer

an=n−1na_n=\dfrac{n-1}{n}

(iii)

3,8,13,18,…3, 8, 13, 18, \ldots

  1. Difference between successive terms: 8−3=58-3=5, 13−8=513-8=5, 18−13=518-13=5.

  2. So the terms are multiples of 55 shifted: 5(1)−2=35(1)-2=3, 5(2)−2=85(2)-2=8, 5(3)−2=135(3)-2=13.

  3. Hence an=5n−2a_n=5n-2.

Answer

an=5n−2a_n=5n-2

Q4. Find the indicated terms of the sequences whose nthn^{th} terms are given …

Find the indicated terms of the sequences whose nthn^{th} terms are given by

(i)

an=5nn+2a_n = \dfrac{5n}{n+2} ; a6a_6 and a13a_{13}

  1. a6=5(6)6+2=308=154a_6=\dfrac{5(6)}{6+2}=\dfrac{30}{8}=\dfrac{15}{4}

  2. a13=5(13)13+2=6515=133a_{13}=\dfrac{5(13)}{13+2}=\dfrac{65}{15}=\dfrac{13}{3}

Answer

a6=154,a13=133a_6=\dfrac{15}{4},\quad a_{13}=\dfrac{13}{3}

(ii)

an=−(n2−4)a_n = -(n^2 - 4); a4a_4 and a11a_{11}

  1. a4=−(42−4)=−(16−4)=−12a_4=-(4^2-4)=-(16-4)=-12

  2. a11=−(112−4)=−(121−4)=−117a_{11}=-(11^2-4)=-(121-4)=-117

Answer

a4=−12,a11=−117a_4=-12,\quad a_{11}=-117

Q5. Find a8a_8 and a15a_{15} whose nthn^{th} term is an={n2−1n+3; n is even, n∈Nn22n+1; n is odd, n∈Na_n = \begin{cases} \dfrac{n^2-1}{n+3} & ;\ n \text{ is even},\ n \in \mathbb{N} \\[2ex] \dfrac{n^2}{2n+1} & ;\ n \text{ is odd},\ n \in \mathbb{N} \end{cases}

  1. 88 is even, so use an=n2−1n+3a_n=\dfrac{n^2-1}{n+3}.

  2. a8=82−18+3=6311a_8=\dfrac{8^2-1}{8+3}=\dfrac{63}{11}

  3. 1515 is odd, so use an=n22n+1a_n=\dfrac{n^2}{2n+1}.

  4. a15=1522(15)+1=22531a_{15}=\dfrac{15^2}{2(15)+1}=\dfrac{225}{31}

Answer

a8=6311,a15=22531a_8=\dfrac{63}{11},\quad a_{15}=\dfrac{225}{31}

Q6. If a1=1a_1 = 1 , a2=1a_2 = 1 and an=2an−1+an−2a_n = 2a_{n-1} + a_{n-2} , n≥3n \geq 3 , n∈Nn \in \mathbb{N} , then …

If a1=1a_1 = 1, a2=1a_2 = 1 and an=2an−1+an−2a_n = 2a_{n-1} + a_{n-2}, n≥3n \geq 3, n∈Nn \in \mathbb{N}, then find the first six terms of the sequence.

  1. Given a1=1a_1=1, a2=1a_2=1.

  2. a3=2a2+a1=2(1)+1=3a_3=2a_2+a_1=2(1)+1=3

  3. a4=2a3+a2=2(3)+1=7a_4=2a_3+a_2=2(3)+1=7

  4. a5=2a4+a3=2(7)+3=17a_5=2a_4+a_3=2(7)+3=17

  5. a6=2a5+a4=2(17)+7=41a_6=2a_5+a_4=2(17)+7=41

Answer

1, 1, 3, 7, 17, 411,\ 1,\ 3,\ 7,\ 17,\ 41

Quick check

  1. If A=265A = 2^{65} and B=264+263+262+⋯+20B = 2^{64} + 2^{63} + 2^{62} + \cdots + 2^{0} which of the following is true?

    1. (1) BB is 2642^{64} more than AA
    2. (2) AA and BB are equal
    3. (3) BB is larger than AA by 11
    4. (4) AA is larger than BB by 11
    Show answer

    Answer: (4) AA is larger than BB by 11

    1. B=20+21+⋯+264B = 2^0 + 2^1 + \cdots + 2^{64} is a G.P. with a=1a = 1, r=2r = 2, n=65n = 65.

    2. B=1(265−1)2−1=265−1B = \dfrac{1(2^{65} - 1)}{2 - 1} = 2^{65} - 1.

    3. A−B=265−(265−1)=1A - B = 2^{65} - (2^{65} - 1) = 1.

    Question 1 of 5

  2. The sum of the exponents of the prime factors in the prime factorization of 17291729 is

    1. (1) 11
    2. (2) 22
    3. (3) 33
    4. (4) 44
    Show answer

    Answer: (3) 33

    1. 1729=7×13×19=71×131×1911729 = 7 \times 13 \times 19 = 7^1 \times 13^1 \times 19^1.

    2. Sum of exponents =1+1+1=3= 1 + 1 + 1 = 3.

    Question 2 of 5

  3. If 66 times of 6th6^{\text{th}} term of an A.P. is equal to 77 times the 7th7^{\text{th}} term, then the 13th13^{\text{th}} term of the A.P. is

    1. (1) 00
    2. (2) 66
    3. (3) 77
    4. (4) 1313
    Show answer

    Answer: (1) 00

    1. 6t6=7t7⇒6(a+5d)=7(a+6d)6t_6 = 7t_7 \Rightarrow 6(a+5d) = 7(a+6d).

    2. 6a+30d=7a+42d⇒a+12d=06a + 30d = 7a + 42d \Rightarrow a + 12d = 0.

    3. t13=a+12d=0t_{13} = a + 12d = 0.

    Question 3 of 5

  4. An A.P. consists of 3131 terms. If its 16th16^{\text{th}} term is mm, then the sum of all the terms of this A.P. is

    1. (1) 16 m16\,m
    2. (2) 62 m62\,m
    3. (3) 31 m31\,m
    4. (4) 312 m\dfrac{31}{2}\,m
    Show answer

    Answer: (3) 31 m31\,m

    1. t16=a+15d=mt_{16} = a + 15d = m.

    2. S31=312 [2a+30d]=312×2(a+15d)S_{31} = \dfrac{31}{2}\,[2a + 30d] = \dfrac{31}{2} \times 2(a + 15d).

    3. =31(a+15d)=31m= 31(a+15d) = 31m.

    Question 4 of 5

  5. Using Euclid's division lemma, if the cube of any positive integer is divided by 99 then the possible remainders are

    1. (1) 0,1,80, 1, 8
    2. (2) 1,4,81, 4, 8
    3. (3) 0,1,30, 1, 3
    4. (4) 1,3,51, 3, 5
    Show answer

    Answer: (1) 0,1,80, 1, 8

    1. Any positive integer is of the form 3q3q, 3q+13q+1 or 3q+23q+2.

    2. (3q)3=27q3=9(3q3)⇒(3q)^3 = 27q^3 = 9(3q^3) \Rightarrow remainder 00.

    3. (3q+1)3=27q3+27q2+9q+1=9(3q3+3q2+q)+1⇒(3q+1)^3 = 27q^3+27q^2+9q+1 = 9(3q^3+3q^2+q)+1 \Rightarrow remainder 11.

    4. (3q+2)3=27q3+54q2+36q+8=9(3q3+6q2+4q)+8⇒(3q+2)^3 = 27q^3+54q^2+36q+8 = 9(3q^3+6q^2+4q)+8 \Rightarrow remainder 88.

    Question 5 of 5
Chapter 2 formulas
  • Euclid's division lemma

    a=bq+r, 0≤r<∣b∣a = bq + r,\ 0 \le r < |b|

  • Fundamental theorem of arithmetic

    N=p1a1p2a2p3a3⋯pkak (unique except for order)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, …a,\ a+d,\ a+2d,\ a+3d,\ \ldots

  • n-th term of an A.P.

    tn=a+(n−1)dt_n = a + (n-1)d

  • Sum to first n terms of an A.P.

    Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}\left[2a + (n-1)d\right]

  • Sum of an A.P. when the last term l is given

    Sn=n2[a+l]S_n = \frac{n}{2}\left[a + l\right]

  • Geometric Progression

    a, ar, ar2, …, arn−1a,\ ar,\ ar^2,\ \ldots,\ ar^{n-1}

  • n-th term of a G.P.

    tn=arn−1t_n = ar^{n-1}

  • Sum to first n terms of a G.P. (r ≠ 1)

    Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1}

  • Sum to first n terms of a G.P. when r = 1

    Sn=naS_n = na

  • Sum to infinite terms of a G.P. (−1 < r < 1)

    S∞=a1−rS_\infty = \frac{a}{1 - r}

  • Sum of first n natural numbers

    1+2+3+⋯+n=n(n+1)21 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}

  • Sum of squares of first n natural numbers

    12+22+32+⋯+n2=n(n+1)(2n+1)61^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}

  • Sum of cubes of first n natural numbers

    13+23+33+⋯+n3=[n(n+1)2]21^3 + 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+⋯+(2n−1)=n21 + 3 + 5 + \cdots + (2n-1) = n^2

Answers checked against the Samacheer Kalvi textbook · Updated

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