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Samacheer Kalvi 10th Maths Exercise 2.8 Solutions

10th Maths Exercise 2.8 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 glance12 answers · full steps below
  1. Q1 (i)Sn=258[1−(−35)n]S_n=\dfrac{25}{8}\left[1-\left(\dfrac{-3}{5}\right)^{n}\right]
  2. Q1 (ii)Sn=10243[1−(14)n]S_n=\dfrac{1024}{3}\left[1-\left(\dfrac{1}{4}\right)^{n}\right]
  3. Q2S6=1820S_6=1820
  4. Q3First term a=12a=12
  5. Q4 (i)S∞=272S_\infty=\dfrac{27}{2}
  6. Q4 (ii)S∞=63S_\infty=63
  7. Q5r=14r=\dfrac{1}{4}
  8. Q6 (i)Sn=49[n−19(1−110n)]S_n=\dfrac{4}{9}\left[n-\dfrac{1}{9}\left(1-\dfrac{1}{10^{n}}\right)\right]
  9. Q6 (ii)Sn=127(10 n+1−10−9n)S_n=\dfrac{1}{27}\left(10^{\,n+1}-10-9n\right)
  10. Q7S10=3069S_{10}=3069
  11. Q8₹1,74,760
  12. Q90.123‾=413330.\overline{123}=\dfrac{41}{333}

Q1. Find the sum of first nn terms of the G.P.

(i)

5,−3,95,−2725,…5, -3, \dfrac{9}{5}, -\dfrac{27}{25}, \ldots

  1. a=5a=5, r=−35r=\dfrac{-3}{5}.

  2. Sn=a(1−rn)1−rS_n=\dfrac{a(1-r^n)}{1-r} (since r<1r<1).

  3. Sn=5(1−(−35)n)1−(−35)=5(1−(−35)n)85S_n=\dfrac{5\left(1-\left(\dfrac{-3}{5}\right)^n\right)}{1-\left(\dfrac{-3}{5}\right)}=\dfrac{5\left(1-\left(\dfrac{-3}{5}\right)^n\right)}{\dfrac{8}{5}}

  4. Sn=258[1−(−35)n]S_n=\dfrac{25}{8}\left[1-\left(\dfrac{-3}{5}\right)^n\right]

Answer

Sn=258[1−(−35)n]S_n=\dfrac{25}{8}\left[1-\left(\dfrac{-3}{5}\right)^{n}\right]

(ii)

256,64,16,…256, 64, 16, \ldots

  1. a=256a=256, r=64256=14r=\dfrac{64}{256}=\dfrac{1}{4}.

  2. Sn=a(1−rn)1−r=256(1−(14)n)1−14S_n=\dfrac{a(1-r^n)}{1-r}=\dfrac{256\left(1-\left(\dfrac{1}{4}\right)^n\right)}{1-\dfrac{1}{4}}

  3. =256×43[1−(14)n]=256\times\dfrac{4}{3}\left[1-\left(\dfrac{1}{4}\right)^n\right]

  4. Sn=10243[1−(14)n]S_n=\dfrac{1024}{3}\left[1-\left(\dfrac{1}{4}\right)^n\right]

Answer

Sn=10243[1−(14)n]S_n=\dfrac{1024}{3}\left[1-\left(\dfrac{1}{4}\right)^{n}\right]

Q2. Find the sum of first six terms of the G.P. 5,15,45,…5, 15, 45, \ldots

  1. a=5a=5, r=155=3>1r=\dfrac{15}{5}=3>1, n=6n=6.

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

  3. S6=5(36−1)3−1=5(729−1)2S_6=\dfrac{5(3^6-1)}{3-1}=\dfrac{5(729-1)}{2}

  4. =5×7282=1820=\dfrac{5\times 728}{2}=1820

Answer

S6=1820S_6=1820

Q3. Find the first term of the G.P. whose common ratio 55 and …

Find the first term of the G.P. whose common ratio 55 and whose sum to first 66 terms is 4687246872.

  1. r=5r=5, S6=46872S_6=46872.

  2. S6=a(r6−1)r−1=46872S_6=\dfrac{a(r^6-1)}{r-1}=46872

  3. a(56−1)5−1=46872\dfrac{a(5^6-1)}{5-1}=46872

  4. a(15625−1)4=46872⇒a×3906=46872\dfrac{a(15625-1)}{4}=46872\Rightarrow a\times 3906=46872

  5. a=468723906=12a=\dfrac{46872}{3906}=12

Answer

First term a=12a=12

Q4. Find the sum to infinity of

(i)

9+3+1+⋯9 + 3 + 1 + \cdots

  1. a=9a=9, r=39=13r=\dfrac{3}{9}=\dfrac{1}{3} and −1<r<1-1<r<1.

  2. S∞=a1−r=91−13S_\infty=\dfrac{a}{1-r}=\dfrac{9}{1-\dfrac{1}{3}}

  3. =923=272=\dfrac{9}{\dfrac{2}{3}}=\dfrac{27}{2}

Answer

S∞=272S_\infty=\dfrac{27}{2}

(ii)

21+14+283+⋯21 + 14 + \dfrac{28}{3} + \cdots

  1. a=21a=21, r=1421=23r=\dfrac{14}{21}=\dfrac{2}{3} and −1<r<1-1<r<1.

  2. S∞=a1−r=211−23S_\infty=\dfrac{a}{1-r}=\dfrac{21}{1-\dfrac{2}{3}}

  3. =2113=63=\dfrac{21}{\dfrac{1}{3}}=63

Answer

S∞=63S_\infty=63

Q5. If the first term of an infinite G.P. is 88 and its …

If the first term of an infinite G.P. is 88 and its sum to infinity is 323\dfrac{32}{3} then find the common ratio.

  1. a=8a=8, S∞=323S_\infty=\dfrac{32}{3}.

  2. a1−r=323⇒81−r=323\dfrac{a}{1-r}=\dfrac{32}{3}\Rightarrow \dfrac{8}{1-r}=\dfrac{32}{3}

  3. 1−r=8×332=341-r=\dfrac{8\times 3}{32}=\dfrac{3}{4}

  4. r=1−34=14r=1-\dfrac{3}{4}=\dfrac{1}{4}

Answer

r=14r=\dfrac{1}{4}

Q6. Find the sum to nn terms of the series

(i)

0.4+0.44+0.444+⋯0.4 + 0.44 + 0.444 + \cdots to nn terms

Show steps (6)

  1. 0.4+0.44+0.444+⋯=4(0.1+0.11+0.111+⋯ )0.4+0.44+0.444+\cdots=4(0.1+0.11+0.111+\cdots)

  2. =49(0.9+0.99+0.999+⋯ )=\dfrac{4}{9}(0.9+0.99+0.999+\cdots) to nn terms

  3. =49[(1−0.1)+(1−0.01)+(1−0.001)+⋯]=\dfrac{4}{9}\big[(1-0.1)+(1-0.01)+(1-0.001)+\cdots\big]

  4. =49[n−(0.1+0.01+0.001+⋯n terms)]=\dfrac{4}{9}\big[n-(0.1+0.01+0.001+\cdots n\text{ terms})\big]

  5. The bracketed part is a G.P. with a=110a=\dfrac{1}{10}, r=110r=\dfrac{1}{10}:

  6. =49[n−110(1−110n)1−110]=49[n−19(1−110n)]=\dfrac{4}{9}\left[n-\dfrac{\dfrac{1}{10}\left(1-\dfrac{1}{10^n}\right)}{1-\dfrac{1}{10}}\right]=\dfrac{4}{9}\left[n-\dfrac{1}{9}\left(1-\dfrac{1}{10^n}\right)\right]

Answer

Sn=49[n−19(1−110n)]S_n=\dfrac{4}{9}\left[n-\dfrac{1}{9}\left(1-\dfrac{1}{10^{n}}\right)\right]

(ii)

3+33+333+⋯3 + 33 + 333 + \cdots to nn terms

Show steps (6)

  1. 3+33+333+⋯=3(1+11+111+⋯ )3+33+333+\cdots=3(1+11+111+\cdots)

  2. =39(9+99+999+⋯ )=\dfrac{3}{9}(9+99+999+\cdots) to nn terms

  3. =13[(10−1)+(100−1)+(1000−1)+⋯]=\dfrac{1}{3}\big[(10-1)+(100-1)+(1000-1)+\cdots\big]

  4. =13[(10+102+103+⋯n terms)−n]=\dfrac{1}{3}\big[(10+10^2+10^3+\cdots n\text{ terms})-n\big]

  5. =13[10(10n−1)10−1−n]=\dfrac{1}{3}\left[\dfrac{10(10^n-1)}{10-1}-n\right]

  6. =10(10n−1)27−n3=10n+1−10−9n27=\dfrac{10(10^n-1)}{27}-\dfrac{n}{3}=\dfrac{10^{n+1}-10-9n}{27}

Answer

Sn=127(10 n+1−10−9n)S_n=\dfrac{1}{27}\left(10^{\,n+1}-10-9n\right)

Q7. Find the sum of the Geometric series 3+6+12+⋯+15363 + 6 + 12 + \cdots + 1536 .

  1. a=3a=3, r=63=2r=\dfrac{6}{3}=2, last term tn=1536t_n=1536.

  2. arn−1=1536⇒3×2n−1=1536ar^{n-1}=1536\Rightarrow 3\times 2^{n-1}=1536

  3. 2n−1=512=29⇒n=102^{n-1}=512=2^9\Rightarrow n=10

  4. S10=a(rn−1)r−1=3(210−1)2−1S_{10}=\dfrac{a(r^n-1)}{r-1}=\dfrac{3(2^{10}-1)}{2-1}

  5. =3(1024−1)=3×1023=3069=3(1024-1)=3\times 1023=3069

Answer

S10=3069S_{10}=3069

Q8. Kumar writes a letter to four of his friends. He asks each …

Kumar writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with the instruction that they continue the process similarly. Assuming that the process is unaltered and it costs ₹2 to mail one letter, find the amount spent on postage when 8th8^{\text{th}} set of letters is mailed.

  1. Letters mailed: 1st set =4=4, 2nd set =42=4^2, 3rd set =43,…=4^3,\ldots

  2. This is a G.P. with a=4a=4, r=4r=4, n=8n=8.

  3. Total letters =4(48−1)4−1=4(65536−1)3=\dfrac{4(4^8-1)}{4-1}=\dfrac{4(65536-1)}{3}

  4. =4×655353=87380=\dfrac{4\times 65535}{3}=87380

  5. Postage =87380×2=₹1,74,760=87380\times 2=₹1,74,760

Answer

₹1,74,760

Q9. Find the rational form of the number 0.123‾0.\overline{123} .

  1. 0.123‾=0.123+0.000123+0.000000123+⋯0.\overline{123}=0.123+0.000123+0.000000123+\cdots

  2. This is a G.P. with a=0.123a=0.123 and r=0.0001230.123=0.001r=\dfrac{0.000123}{0.123}=0.001, and −1<r<1-1<r<1.

  3. S∞=a1−r=0.1231−0.001=0.1230.999S_\infty=\dfrac{a}{1-r}=\dfrac{0.123}{1-0.001}=\dfrac{0.123}{0.999}

  4. =123999=41333=\dfrac{123}{999}=\dfrac{41}{333}

Answer

0.123‾=413330.\overline{123}=\dfrac{41}{333}

Q10. If Sn=(x+y)+(x2+xy+y2)+(x3+x2y+xy2+y3)+⋯nS_n = (x+y) + (x^2 + xy + y^2) + (x^3 + x^2y + xy^2 + y^3) + \cdots n terms then prove that (x−y)Sn=[x2(xn−1)x−1−y2(yn−1)y−1](x-y)S_n = \left[\dfrac{x^2(x^n - 1)}{x-1} - \dfrac{y^2(y^n - 1)}{y-1}\right]

Show steps (9)

  1. Multiply each term by (x−y)(x-y) and use (x−y)(xk+xk−1y+⋯+yk)=xk+1−yk+1(x-y)(x^{k}+x^{k-1}y+\cdots+y^{k})=x^{k+1}-y^{k+1}.

  2. (x−y)(x+y)=x2−y2(x-y)(x+y)=x^2-y^2

  3. (x−y)(x2+xy+y2)=x3−y3(x-y)(x^2+xy+y^2)=x^3-y^3

  4. (x−y)(x3+x2y+xy2+y3)=x4−y4(x-y)(x^3+x^2y+xy^2+y^3)=x^4-y^4, and so on.

  5. So (x−y)Sn=(x2−y2)+(x3−y3)+⋯+(xn+1−yn+1)(x-y)S_n=(x^2-y^2)+(x^3-y^3)+\cdots+(x^{n+1}-y^{n+1})

  6. =(x2+x3+⋯+xn+1)−(y2+y3+⋯+yn+1)=(x^2+x^3+\cdots+x^{n+1})-(y^2+y^3+\cdots+y^{n+1})

  7. Each bracket is a G.P. of nn terms; first with a=x2a=x^2, r=xr=x; second with a=y2a=y^2, r=yr=y.

  8. =x2(xn−1)x−1−y2(yn−1)y−1=\dfrac{x^2(x^n-1)}{x-1}-\dfrac{y^2(y^n-1)}{y-1}

  9. Hence (x−y)Sn=[x2(xn−1)x−1−y2(yn−1)y−1](x-y)S_n=\left[\dfrac{x^2(x^n-1)}{x-1}-\dfrac{y^2(y^n-1)}{y-1}\right].

Answer

(x−y)Sn=[x2(xn−1)x−1−y2(yn−1)y−1](x-y)S_n=\left[\dfrac{x^{2}(x^{n}-1)}{x-1}-\dfrac{y^{2}(y^{n}-1)}{y-1}\right] — proved

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

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