Search

Samacheer Kalvi 10th Maths Exercise 2.7 Solutions

10th Maths Exercise 2.7 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 glance19 answers · full steps below
  1. Q1 (i)It is a G.P. with common ratio r=3r=3
  2. Q1 (ii)It is not a G.P.
  3. Q1 (iii)It is a G.P. with common ratio r=110r=\dfrac{1}{10}
  4. Q1 (iv)It is a G.P. with common ratio r=12r=\dfrac{1}{2}
  5. Q1 (v)It is a G.P. with common ratio r=−5r=-5
  6. Q1 (vi)It is not a G.P.
  7. Q1 (vii)It is a G.P. with common ratio r=14r=\dfrac{1}{4}
  8. Q2 (i)6, 18, 546,\ 18,\ 54
  9. Q2 (ii)2, 2, 22\sqrt{2},\ 2,\ 2\sqrt{2}
  10. Q2 (iii)1000, 400, 1601000,\ 400,\ 160
  11. Q3t7=1t_7=1
  12. Q4x=−18x=-18
  13. Q5 (i)1212 terms
  14. Q5 (ii)77 terms
  15. Q6t12=885735t_{12}=885735
  16. Q7t10=3072t_{10}=3072
  17. Q92, 3, 922,\ 3,\ \dfrac{9}{2} (or 92, 3, 2\dfrac{9}{2},\ 3,\ 2)
  18. Q10₹76,576.8976{,}576.89
  19. Q11Offer A: ₹23,820.3223{,}820.32; Offer B: ₹24,039.9924{,}039.99 (Offer B is better)

Q1. Which of the following sequences are in G.P.?

(i)

3,9,27,81,…3, 9, 27, 81,\dots

  1. t2t1=93=3\dfrac{t_2}{t_1}=\dfrac{9}{3}=3

  2. t3t2=279=3\dfrac{t_3}{t_2}=\dfrac{27}{9}=3, t4t3=8127=3\dfrac{t_4}{t_3}=\dfrac{81}{27}=3

  3. All ratios are equal.

Answer

It is a G.P. with common ratio r=3r=3

(ii)

4,44,444,4444,...4, 44, 444, 4444,...

  1. t2t1=444=11\dfrac{t_2}{t_1}=\dfrac{44}{4}=11

  2. t3t2=44444=11111\dfrac{t_3}{t_2}=\dfrac{444}{44}=\dfrac{111}{11}

  3. The ratios are not equal.

Answer

It is not a G.P.

(iii)

0.5,0.05,0.005,…0.5, 0.05, 0.005,\dots

  1. t2t1=0.050.5=110\dfrac{t_2}{t_1}=\dfrac{0.05}{0.5}=\dfrac{1}{10}

  2. t3t2=0.0050.05=110\dfrac{t_3}{t_2}=\dfrac{0.005}{0.05}=\dfrac{1}{10}

  3. The ratios are equal.

Answer

It is a G.P. with common ratio r=110r=\dfrac{1}{10}

(iv)

13,16,112,...\dfrac{1}{3}, \dfrac{1}{6}, \dfrac{1}{12},...

  1. t2t1=1/61/3=12\dfrac{t_2}{t_1}=\dfrac{1/6}{1/3}=\dfrac{1}{2}

  2. t3t2=1/121/6=12\dfrac{t_3}{t_2}=\dfrac{1/12}{1/6}=\dfrac{1}{2}

  3. The ratios are equal.

Answer

It is a G.P. with common ratio r=12r=\dfrac{1}{2}

(v)

1,−5,25,−125,…1, -5, 25, -125,\dots

  1. t2t1=−51=−5\dfrac{t_2}{t_1}=\dfrac{-5}{1}=-5

  2. t3t2=25−5=−5\dfrac{t_3}{t_2}=\dfrac{25}{-5}=-5, t4t3=−12525=−5\dfrac{t_4}{t_3}=\dfrac{-125}{25}=-5

  3. All ratios are equal.

Answer

It is a G.P. with common ratio r=−5r=-5

(vi)

120,60,30,18,…120, 60, 30, 18,\dots

  1. t2t1=60120=12\dfrac{t_2}{t_1}=\dfrac{60}{120}=\dfrac{1}{2}

  2. t3t2=3060=12\dfrac{t_3}{t_2}=\dfrac{30}{60}=\dfrac{1}{2}

  3. t4t3=1830=35\dfrac{t_4}{t_3}=\dfrac{18}{30}=\dfrac{3}{5}

  4. The ratios are not all equal.

Answer

It is not a G.P.

(vii)

16,4,1,14,...16, 4, 1, \dfrac{1}{4},...

  1. t2t1=416=14\dfrac{t_2}{t_1}=\dfrac{4}{16}=\dfrac{1}{4}

  2. t3t2=14\dfrac{t_3}{t_2}=\dfrac{1}{4}, t4t3=1/41=14\dfrac{t_4}{t_3}=\dfrac{1/4}{1}=\dfrac{1}{4}

  3. All ratios are equal.

Answer

It is a G.P. with common ratio r=14r=\dfrac{1}{4}

Q2. Write the first three terms of the G.P. whose first term and …

Write the first three terms of the G.P. whose first term and the common ratio are given below.

(i)

a=6a = 6, r=3r = 3

  1. General form: a, ar, ar2a,\ ar,\ ar^2

  2. a=6a=6

  3. ar=6×3=18ar=6\times 3=18

  4. ar2=6×32=54ar^2=6\times 3^2=54

Answer

6, 18, 546,\ 18,\ 54

(ii)

a=2a = \sqrt{2}, r=2r = \sqrt{2}

  1. General form: a, ar, ar2a,\ ar,\ ar^2

  2. a=2a=\sqrt{2}

  3. ar=2×2=2ar=\sqrt{2}\times\sqrt{2}=2

  4. ar2=2×2=22ar^2=\sqrt{2}\times 2=2\sqrt{2}

Answer

2, 2, 22\sqrt{2},\ 2,\ 2\sqrt{2}

(iii)

a=1000a = 1000, r=25r = \dfrac{2}{5}

  1. General form: a, ar, ar2a,\ ar,\ ar^2

  2. a=1000a=1000

  3. ar=1000×25=400ar=1000\times\dfrac{2}{5}=400

  4. ar2=1000×425=160ar^2=1000\times\dfrac{4}{25}=160

Answer

1000, 400, 1601000,\ 400,\ 160

Q3. In a G.P. 729,243,81,…729, 243, 81,\dots find t7t_7 .

  1. a=729a=729, r=243729=13r=\dfrac{243}{729}=\dfrac{1}{3}

  2. tn=arn−1t_n=ar^{n-1}

  3. t7=729×(13)6t_7=729\times\left(\dfrac{1}{3}\right)^{6}

  4. =729729=1=\dfrac{729}{729}=1

Answer

t7=1t_7=1

Q4. Find xx so that x+6x + 6 , x+12x + 12 and x+15x + 15 are consecutive terms …

Find xx so that x+6x + 6, x+12x + 12 and x+15x + 15 are consecutive terms of a Geometric Progression.

  1. For three terms in G.P., b2=acb^2=ac

  2. (x+12)2=(x+6)(x+15)(x+12)^2=(x+6)(x+15)

  3. x2+24x+144=x2+21x+90x^2+24x+144=x^2+21x+90

  4. 3x=−543x=-54

  5. x=−18x=-18

Answer

x=−18x=-18

Q5. Find the number of terms in the following G.P.

(i)

4,8,16,…,81924, 8, 16,\dots, 8192?

  1. a=4a=4, r=84=2r=\dfrac{8}{4}=2, tn=8192t_n=8192

  2. arn−1=8192ar^{n-1}=8192

  3. 4×2 n−1=81924\times 2^{\,n-1}=8192

  4. 2 n−1=2048=2112^{\,n-1}=2048=2^{11}

  5. n−1=11⇒n=12n-1=11\Rightarrow n=12

Answer

1212 terms

(ii)

13,19,127,...,12187\dfrac{1}{3}, \dfrac{1}{9}, \dfrac{1}{27},..., \dfrac{1}{2187}

  1. a=13a=\dfrac{1}{3}, r=1/91/3=13r=\dfrac{1/9}{1/3}=\dfrac{1}{3}, tn=12187t_n=\dfrac{1}{2187}

  2. 13×(13)n−1=12187\dfrac{1}{3}\times\left(\dfrac{1}{3}\right)^{n-1}=\dfrac{1}{2187}

  3. (13)n=137\left(\dfrac{1}{3}\right)^{n}=\dfrac{1}{3^{7}}

  4. n=7n=7

Answer

77 terms

Q6. In a G.P. the 9th9^{th} term is 3280532805 and 6th6^{th} term is …

In a G.P. the 9th9^{th} term is 3280532805 and 6th6^{th} term is 12151215. Find the 12th12^{th} term.

Show steps (6)

  1. t9=ar8=32805t_9=ar^8=32805 …(1)

  2. t6=ar5=1215t_6=ar^5=1215 …(2)

  3. Dividing (1) by (2): r3=328051215=27r^3=\dfrac{32805}{1215}=27

  4. r=3r=3

  5. From (2): a×35=1215⇒a=5a\times 3^5=1215\Rightarrow a=5

  6. t12=ar11=5×311=5×177147t_{12}=ar^{11}=5\times 3^{11}=5\times 177147

Answer

t12=885735t_{12}=885735

Q7. Find the 10th10^{th} term of a G.P. whose 8th8^{th} term is 768768 …

Find the 10th10^{th} term of a G.P. whose 8th8^{th} term is 768768 and the common ratio is 22.

  1. t8=ar7=768t_8=ar^7=768, r=2r=2

  2. t10=ar9=ar7×r2t_{10}=ar^9=ar^7\times r^2

  3. =768×22=768\times 2^2

  4. =768×4=3072=768\times 4=3072

Answer

t10=3072t_{10}=3072

Q8. If aa , bb , cc are in A.P. then show that …

If aa, bb, cc are in A.P. then show that 3a,3b,3c3^a, 3^b, 3^c are in G.P.

  1. a,b,ca,b,c are in A.P. ⇒2b=a+c\Rightarrow 2b=a+c

  2. For 3a,3b,3c3^a,3^b,3^c to be in G.P. we need (3b)2=3a×3c(3^b)^2=3^a\times 3^c

  3. (3b)2=32b(3^b)^2=3^{2b}

  4. 3a×3c=3a+c=32b3^a\times 3^c=3^{a+c}=3^{2b} (since a+c=2ba+c=2b)

  5. So (3b)2=3a×3c(3^b)^2=3^a\times 3^c

Answer

Since (3b)2=3a⋅3c(3^b)^2=3^a\cdot 3^c, the numbers 3a,3b,3c3^a,3^b,3^c are in G.P.

Q9. In a G.P. the product of three consecutive terms is 2727 and …

In a G.P. the product of three consecutive terms is 2727 and the sum of the product of two terms taken at a time is 572\dfrac{57}{2}. Find the three terms.

Show steps (9)

  1. Let the three terms be ar, a, ar\dfrac{a}{r},\ a,\ ar

  2. Product =a3=27⇒a=3=a^3=27\Rightarrow a=3

  3. Sum of products two at a time:

  4. ar⋅a+a⋅ar+ar⋅ar=572\dfrac{a}{r}\cdot a+a\cdot ar+\dfrac{a}{r}\cdot ar=\dfrac{57}{2}

  5. 9(1r+r+1)=5729\left(\dfrac{1}{r}+r+1\right)=\dfrac{57}{2}

  6. 1r+r=196−1=136\dfrac{1}{r}+r=\dfrac{19}{6}-1=\dfrac{13}{6}

  7. 6r2−13r+6=0⇒(2r−3)(3r−2)=06r^2-13r+6=0\Rightarrow (2r-3)(3r-2)=0

  8. r=32r=\dfrac{3}{2} or r=23r=\dfrac{2}{3}

  9. If a=3, r=32a=3,\ r=\dfrac{3}{2}: terms are 2, 3, 922,\ 3,\ \dfrac{9}{2}

Answer

2, 3, 922,\ 3,\ \dfrac{9}{2} (or 92, 3, 2\dfrac{9}{2},\ 3,\ 2)

Q10. A man joined a company as Assistant Manager. The company gave him …

A man joined a company as Assistant Manager. The company gave him a starting salary of ₹60,00060,000 and agreed to increase his salary 5%5\% annually. What will be his salary after 55 years?

  1. Salaries form a G.P. with a=60000a=60000, r=1+5100=2120r=1+\dfrac{5}{100}=\dfrac{21}{20}

  2. Salary after 5 years =t6=ar5=t_6=ar^5

  3. =60000×(2120)5=60000\times\left(\dfrac{21}{20}\right)^{5}

  4. =60000×1.2762815625=60000\times 1.2762815625

  5. =76576.89=76576.89

Answer

₹76,576.8976{,}576.89

Q11. Sivamani is attending an interview for a job and the company gave …

Sivamani is attending an interview for a job and the company gave two offers to him. Offer A: ₹20,00020,000 to start with followed by a guaranteed annual increase of 6%6\% for the first 55 years. Offer B: ₹22,00022,000 to start with followed by a guaranteed annual increase of 3%3\% for the first 55 years. What is his salary in the 4th4^{th} year with respect to the offers AA and BB?

  1. Offer A: a=20000a=20000, r=1+6100=5350r=1+\dfrac{6}{100}=\dfrac{53}{50}

  2. t4=ar3=20000×(1.06)3=20000×1.191016=23820.32t_4=ar^3=20000\times(1.06)^3=20000\times1.191016=23820.32

  3. Offer B: a=22000a=22000, r=1+3100=103100r=1+\dfrac{3}{100}=\dfrac{103}{100}

  4. t4=ar3=22000×(1.03)3=22000×1.092727=24039.99t_4=ar^3=22000\times(1.03)^3=22000\times1.092727=24039.99

Answer

Offer A: ₹23,820.3223{,}820.32; Offer B: ₹24,039.9924{,}039.99 (Offer B is better)

Q12. If aa , bb , cc are three consecutive terms of an …

If aa, bb, cc are three consecutive terms of an A.P. and xx, yy, zz are three consecutive terms of a G.P. then prove that xb−c×yc−a×za−b=1x^{b-c} \times y^{c-a} \times z^{a-b} = 1.

Show steps (8)

  1. a,b,ca,b,c in A.P. ⇒2b=a+c\Rightarrow 2b=a+c

  2. x,y,zx,y,z in G.P. ⇒y=xr, z=xr2\Rightarrow y=xr,\ z=xr^2

  3. xb−c×yc−a×za−bx^{b-c}\times y^{c-a}\times z^{a-b}

  4. =xb−c (xr)c−a (xr2)a−b=x^{b-c}\,(xr)^{c-a}\,(xr^2)^{a-b}

  5. =x(b−c)+(c−a)+(a−b)⋅r(c−a)+2(a−b)=x^{(b-c)+(c-a)+(a-b)}\cdot r^{(c-a)+2(a-b)}

  6. =x0⋅r a+c−2b=x^{0}\cdot r^{\,a+c-2b}

  7. Since a+c=2ba+c=2b,  a+c−2b=0\ a+c-2b=0

  8. =1×r0=1=1\times r^{0}=1

Answer

xb−c×yc−a×za−b=1x^{b-c}\times y^{c-a}\times z^{a-b}=1

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.