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Samacheer Kalvi 10th Maths Exercise 2.10 Solutions

10th Maths Exercise 2.10 solutions (Chapter 2 Numbers and Sequences) with step-by-step answers for all 15 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance15 answers · full steps below
  1. Q1
    1. 0≤r<b0 \leq r < b
  2. Q2
    1. 0,1,80, 1, 8
  3. Q3
    1. 22
  4. Q4
    1. 33
  5. Q5
    1. 25202520
  6. Q6
    1. 11
  7. Q7
    1. 1111
  8. Q8
    1. 78817881
  9. Q9
    1. 00
  10. Q10
    1. 31 m31\,m
  11. Q11
    1. 88
  12. Q12
    1. AA is larger than BB by 11
  13. Q13
    1. 127\dfrac{1}{27}
  14. Q14
    1. an Arithmetic Progression
  15. Q15
    1. 1428014280

Q1. Euclid's division lemma states that for positive integers aa and bb …

Euclid's division lemma states that for positive integers aa and bb, there exist unique integers qq and rr such that a=bq+ra = bq + r, where rr must satisfy.

  1. (1) 1<r<b1 < r < b
  2. (2) 0<r<b0 < r < b
  3. (3) 0≤r<b0 \leq r < b
  4. (4) 0<r≤b0 < r \leq b

  1. In a=bq+ra = bq + r, the remainder rr can be zero (when bb divides aa) but must be less than the divisor bb.

  2. So 0≤r<b0 \le r < b.

Answer
  1. 0≤r<b0 \leq r < b

Q2. Using Euclid's division lemma, if the cube of any positive integer is …

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

  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.

Answer
  1. 0,1,80, 1, 8

Q3. If the HCF of 6565 and 117117 is expressible in the form …

If the HCF of 6565 and 117117 is expressible in the form of 65m−11765m - 117, then the value of mm is

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

  1. 117=65×1+52117 = 65 \times 1 + 52, 65=52×1+1365 = 52 \times 1 + 13, 52=13×4+052 = 13 \times 4 + 0.

  2. So HCF(65,117)=13(65,117) = 13.

  3. 65m−117=13⇒65m=130⇒m=265m - 117 = 13 \Rightarrow 65m = 130 \Rightarrow m = 2.

Answer
  1. 22

Q4. The sum of the exponents of the prime factors in the prime …

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

  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.

Answer
  1. 33

Q5. The least number that is divisible by all the numbers from 11 …

The least number that is divisible by all the numbers from 11 to 1010 (both inclusive) is

  1. (1) 20252025
  2. (2) 52205220
  3. (3) 50255025
  4. (4) 25202520

  1. The least such number is the LCM of 1,2,…,101,2,\ldots,10.

  2. LCM=23×32×5×7=2520\text{LCM} = 2^3 \times 3^2 \times 5 \times 7 = 2520.

Answer
  1. 25202520

Q6. 74k≡7^{4k} \equiv _____ (mod 100)(\text{mod } 100)

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

  1. 72=497^2 = 49, so 74=492=24017^4 = 49^2 = 2401.

  2. 2401=24×100+1⇒74≡1(mod100)2401 = 24 \times 100 + 1 \Rightarrow 7^4 \equiv 1 \pmod{100}.

  3. Hence 74k=(74)k≡1k≡1(mod100)7^{4k} = (7^4)^k \equiv 1^k \equiv 1 \pmod{100}.

Answer
  1. 11

Q7. Given F1=1F_1 = 1 , F2=3F_2 = 3 and Fn=Fn−1+Fn−2F_n = F_{n-1} + F_{n-2} then F5F_5 is

  1. (1) 33
  2. (2) 55
  3. (3) 88
  4. (4) 1111

  1. F3=F2+F1=3+1=4F_3 = F_2 + F_1 = 3 + 1 = 4.

  2. F4=F3+F2=4+3=7F_4 = F_3 + F_2 = 4 + 3 = 7.

  3. F5=F4+F3=7+4=11F_5 = F_4 + F_3 = 7 + 4 = 11.

Answer
  1. 1111

Q8. The first term of an arithmetic progression is unity and the common …

The first term of an arithmetic progression is unity and the common difference is 44. Which of the following will be a term of this A.P.

  1. (1) 45514551
  2. (2) 1009110091
  3. (3) 78817881
  4. (4) 1353113531

  1. a=1a = 1, d=4d = 4, so tn=1+(n−1)4=4n−3t_n = 1 + (n-1)4 = 4n - 3.

  2. Every term leaves remainder 11 when divided by 44.

  3. 7881=4×1971−37881 = 4 \times 1971 - 3, so 78817881 is the 1971st1971^{\text{st}} term.

Answer
  1. 78817881

Q9. If 66 times of 6th6^{\text{th}} term of an A.P. is equal to …

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

  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.

Answer
  1. 00

Q10. An A.P. consists of 3131 terms. If its 16th16^{\text{th}} term is mm …

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

  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.

Answer
  1. 31 m31\,m

Q11. In an A.P., the first term is 11 and the common difference …

In an A.P., the first term is 11 and the common difference is 44. How many terms of the A.P. must be taken for their sum to be equal to 120120?

  1. (1) 66
  2. (2) 77
  3. (3) 88
  4. (4) 99

  1. Sn=n2[2(1)+(n−1)4]=n2(4n−2)=n(2n−1)S_n = \dfrac{n}{2}[2(1) + (n-1)4] = \dfrac{n}{2}(4n - 2) = n(2n-1).

  2. n(2n−1)=120⇒2n2−n−120=0n(2n-1) = 120 \Rightarrow 2n^2 - n - 120 = 0.

  3. (n−8)(2n+15)=0⇒n=8(n-8)(2n+15) = 0 \Rightarrow n = 8.

Answer
  1. 88

Q12. 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

  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.

Answer
  1. AA is larger than BB by 11

Q13. The next term of the sequence 316,18,112,118,⋯\dfrac{3}{16}, \dfrac{1}{8}, \dfrac{1}{12}, \dfrac{1}{18}, \cdots is

  1. (1) 124\dfrac{1}{24}
  2. (2) 127\dfrac{1}{27}
  3. (3) 23\dfrac{2}{3}
  4. (4) 181\dfrac{1}{81}

  1. 1/83/16=1624=23\dfrac{1/8}{3/16} = \dfrac{16}{24} = \dfrac{2}{3}, 1/121/8=23\dfrac{1/12}{1/8} = \dfrac{2}{3}, 1/181/12=23\dfrac{1/18}{1/12} = \dfrac{2}{3}.

  2. It is a G.P. with r=23r = \dfrac{2}{3}.

  3. Next term =118×23=127= \dfrac{1}{18} \times \dfrac{2}{3} = \dfrac{1}{27}.

Answer
  1. 127\dfrac{1}{27}

Q14. If the sequence t1,t2,t3,…t_1, t_2, t_3, \ldots are in A.P. then the sequence t6,t12,t18,…t_6, t_{12}, t_{18}, \ldots is

  1. (1) a Geometric Progression
  2. (2) an Arithmetic Progression
  3. (3) neither an Arithmetic Progression nor a Geometric Progression
  4. (4) a constant sequence

  1. Let the A.P. have first term aa and common difference dd.

  2. t6=a+5dt_6 = a+5d, t12=a+11dt_{12} = a+11d, t18=a+17dt_{18} = a+17d.

  3. Difference =6d= 6d, a constant, so the new sequence is an A.P.

Answer
  1. an Arithmetic Progression

Q15. The value of (13+23+33+⋯+153)−(1+2+3+⋯+15)(1^3 + 2^3 + 3^3 + \cdots + 15^3) - (1 + 2 + 3 + \cdots + 15) is

  1. (1) 1440014400
  2. (2) 1420014200
  3. (3) 1428014280
  4. (4) 1452014520

  1. 13+23+⋯+153=[15×162]2=1202=144001^3+2^3+\cdots+15^3 = \left[\dfrac{15 \times 16}{2}\right]^2 = 120^2 = 14400.

  2. 1+2+⋯+15=15×162=1201+2+\cdots+15 = \dfrac{15 \times 16}{2} = 120.

  3. 14400−120=1428014400 - 120 = 14280.

Answer
  1. 1428014280
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

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