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Samacheer Kalvi 10th Maths Exercise 2.6 Solutions

10th Maths Exercise 2.6 solutions (Chapter 2 Numbers and Sequences) with step-by-step answers for all 16 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance14 answers · full steps below
  1. Q1 (i)S40=3240S_{40}=3240
  2. Q1 (ii)S27=999S_{27}=999
  3. Q1 (iii)S14=721S_{14}=721
  4. Q22020 consecutive odd integers
  5. Q3S28=1540S_{28}=1540
  6. Q5S35=12252=612.5S_{35}=\dfrac{1225}{2}=612.5
  7. Q65062550625
  8. Q7168448168448
  9. Q8 (i)₹45,75045{,}750
  10. Q8 (ii)₹57505750 extra
  11. Q92020 months
  12. Q10 (i)4242 bricks in the top most step
  13. Q10 (ii)21302130 bricks
  14. Q126(24a−13b)a+b=144a−78ba+b\dfrac{6(24a-13b)}{a+b}=\dfrac{144a-78b}{a+b}

Q1. Find the sum of the following:

(i)

3,7,11,…3, 7, 11, \ldots up to 4040 terms.

  1. a=3a=3, d=7−3=4d=7-3=4, n=40n=40.

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

  3. S40=402[2(3)+39(4)]S_{40}=\dfrac{40}{2}[2(3)+39(4)]

  4. =20[6+156]=20×162=20[6+156]=20\times 162

  5. =3240=3240

Answer

S40=3240S_{40}=3240

(ii)

102,97,92,…102, 97, 92, \ldots up to 2727 terms.

  1. a=102a=102, d=97−102=−5d=97-102=-5, n=27n=27.

  2. S27=272[2(102)+26(−5)]S_{27}=\dfrac{27}{2}[2(102)+26(-5)]

  3. =272[204−130]=272×74=\dfrac{27}{2}[204-130]=\dfrac{27}{2}\times 74

  4. =27×37=999=27\times 37=999

Answer

S27=999S_{27}=999

(iii)

6+13+20+⋯+976 + 13 + 20 + \cdots + 97

  1. a=6a=6, d=13−6=7d=13-6=7, last term l=97l=97.

  2. n=l−ad+1=97−67+1=13+1=14n=\dfrac{l-a}{d}+1=\dfrac{97-6}{7}+1=13+1=14

  3. Sn=n2[a+l]S_n=\dfrac{n}{2}[a+l]

  4. S14=142[6+97]=7×103S_{14}=\dfrac{14}{2}[6+97]=7\times 103

  5. =721=721

Answer

S14=721S_{14}=721

Q2. How many consecutive odd integers beginning with 55 will sum to 480480 …

How many consecutive odd integers beginning with 55 will sum to 480480?

Show steps (7)

  1. The odd integers are 5,7,9,…5,7,9,\ldots so a=5a=5, d=2d=2.

  2. Sn=n2[2(5)+(n−1)2]=480S_n=\dfrac{n}{2}[2(5)+(n-1)2]=480

  3. n2[10+2n−2]=480\dfrac{n}{2}[10+2n-2]=480

  4. n(n+4)=480n(n+4)=480

  5. n2+4n−480=0n^2+4n-480=0

  6. (n+24)(n−20)=0(n+24)(n-20)=0

  7. n=−24n=-24 is not possible, so n=20n=20.

Answer

2020 consecutive odd integers

Q3. Find the sum of first 2828 terms of an A.P. whose nthn^{\text{th}} …

Find the sum of first 2828 terms of an A.P. whose nthn^{\text{th}} term is 4n−34n - 3.

Show steps (6)

  1. tn=4n−3t_n=4n-3

  2. a=t1=4(1)−3=1a=t_1=4(1)-3=1

  3. t28=4(28)−3=109t_{28}=4(28)-3=109

  4. Sn=n2[a+l]S_n=\dfrac{n}{2}[a+l]

  5. S28=282[1+109]=14×110S_{28}=\dfrac{28}{2}[1+109]=14\times 110

  6. =1540=1540

Answer

S28=1540S_{28}=1540

Q4. The sum of first nn terms of a certain series is given …

The sum of first nn terms of a certain series is given as 2n2−3n2n^2 - 3n. Show that the series is an A.P.

Show steps (7)

  1. Sn=2n2−3nS_n=2n^2-3n

  2. Sn−1=2(n−1)2−3(n−1)=2n2−7n+5S_{n-1}=2(n-1)^2-3(n-1)=2n^2-7n+5

  3. tn=Sn−Sn−1=(2n2−3n)−(2n2−7n+5)t_n=S_n-S_{n-1}=(2n^2-3n)-(2n^2-7n+5)

  4. tn=4n−5t_n=4n-5

  5. tn−1=4(n−1)−5=4n−9t_{n-1}=4(n-1)-5=4n-9

  6. tn−tn−1=(4n−5)−(4n−9)=4t_n-t_{n-1}=(4n-5)-(4n-9)=4, a constant.

  7. Since the difference between consecutive terms is the same, the series is an A.P. with a=−1a=-1, d=4d=4.

Answer

tn=4n−5t_n=4n-5 and tn−tn−1=4t_n-t_{n-1}=4 (constant), so the series is an A.P.

Q5. The 104th104^{\text{th}} term and 4th4^{\text{th}} term of an A.P. are 125125 and …

The 104th104^{\text{th}} term and 4th4^{\text{th}} term of an A.P. are 125125 and 00. Find the sum of first 3535 terms.

Show steps (7)

  1. t104=a+103d=125t_{104}=a+103d=125 …(1)

  2. t4=a+3d=0t_4=a+3d=0 …(2)

  3. (1) −- (2): 100d=125⇒d=54100d=125\Rightarrow d=\dfrac{5}{4}

  4. From (2): a=−3d=−154a=-3d=-\dfrac{15}{4}

  5. S35=352[2(−154)+34(54)]S_{35}=\dfrac{35}{2}\left[2\left(-\dfrac{15}{4}\right)+34\left(\dfrac{5}{4}\right)\right]

  6. =352[−152+852]=352×35=\dfrac{35}{2}\left[-\dfrac{15}{2}+\dfrac{85}{2}\right]=\dfrac{35}{2}\times 35

  7. =12252=612.5=\dfrac{1225}{2}=612.5

Answer

S35=12252=612.5S_{35}=\dfrac{1225}{2}=612.5

Q6. Find the sum of all odd positive integers less than 450450 .

  1. The odd positive integers less than 450450 are 1,3,5,…,4491,3,5,\ldots,449.

  2. a=1a=1, d=2d=2, l=449l=449.

  3. n=449−12+1=224+1=225n=\dfrac{449-1}{2}+1=224+1=225

  4. S225=2252[1+449]=2252×450S_{225}=\dfrac{225}{2}[1+449]=\dfrac{225}{2}\times 450

  5. =225×225=50625=225\times 225=50625

Answer

5062550625

Q7. Find the sum of all natural numbers between 602602 and 902902 which …

Find the sum of all natural numbers between 602602 and 902902 which are not divisible by 44.

Show steps (7)

  1. Natural numbers between 602602 and 902902: 603,604,…,901603,604,\ldots,901.

  2. Here a=603a=603, l=901l=901, d=1d=1, n=901−603+1=299n=901-603+1=299.

  3. Sum of all =2992[603+901]=299×752=224848=\dfrac{299}{2}[603+901]=299\times 752=224848

  4. Numbers divisible by 44: 604,608,…,900604,608,\ldots,900.

  5. n=900−6044+1=74+1=75n=\dfrac{900-604}{4}+1=74+1=75

  6. Their sum =752[604+900]=75×752=56400=\dfrac{75}{2}[604+900]=75\times 752=56400

  7. Required sum =224848−56400=168448=224848-56400=168448

Answer

168448168448

Q8. Raghu wish to buy a laptop. He can buy it by paying …

Raghu wish to buy a laptop. He can buy it by paying ₹40,000₹40{,}000 cash or by giving it in 1010 installments as ₹4800₹4800 in the first month, ₹4750₹4750 in the second month, ₹4700₹4700 in the third month and so on. If he pays the money in this fashion, find

(i)

total amount paid in 1010 installments.

  1. Installments: 4800,4750,4700,…4800,4750,4700,\ldots

  2. a=4800a=4800, d=4750−4800=−50d=4750-4800=-50, n=10n=10.

  3. S10=102[2(4800)+9(−50)]S_{10}=\dfrac{10}{2}[2(4800)+9(-50)]

  4. =5[9600−450]=5×9150=5[9600-450]=5\times 9150

  5. =45750=45750

Answer

₹45,75045{,}750

(ii)

how much extra amount that he has to pay than the cost?

  1. Amount paid in installments == ₹4575045750

  2. Cash price == ₹4000040000

  3. Extra amount =45750−40000=5750=45750-40000=5750

Answer

₹57505750 extra

Q9. A man repays a loan of ₹65,000₹65{,}000 by paying ₹400₹400 in the …

A man repays a loan of ₹65,000₹65{,}000 by paying ₹400₹400 in the first month and then increasing the payment by ₹300₹300 every month. How long will it take for him to clear the loan?

Show steps (8)

  1. Payments: 400,700,1000,…400,700,1000,\ldots so a=400a=400, d=300d=300.

  2. Sn=65000S_n=65000

  3. n2[2(400)+(n−1)300]=65000\dfrac{n}{2}[2(400)+(n-1)300]=65000

  4. n[800+300n−300]=130000n[800+300n-300]=130000

  5. n[300n+500]=130000n[300n+500]=130000

  6. 3n2+5n−1300=03n^2+5n-1300=0

  7. n=−5±25+156006=−5±1256n=\dfrac{-5\pm\sqrt{25+15600}}{6}=\dfrac{-5\pm 125}{6}

  8. n=20n=20 (negative value rejected).

Answer

2020 months

Q10. A brick staircase has a total of 3030 steps. The bottom step …

A brick staircase has a total of 3030 steps. The bottom step requires 100100 bricks. Each successive step requires two bricks less than the previous step.

(i)

How many bricks are required for the top most step?

  1. Bricks in steps from bottom: 100,98,96,…100,98,96,\ldots

  2. a=100a=100, d=−2d=-2, n=30n=30.

  3. t30=a+29d=100+29(−2)t_{30}=a+29d=100+29(-2)

  4. =100−58=42=100-58=42

Answer

4242 bricks in the top most step

(ii)

How many bricks are required to build the stair case?

  1. a=100a=100, l=42l=42, n=30n=30.

  2. S30=302[100+42]S_{30}=\dfrac{30}{2}[100+42]

  3. =15×142=2130=15\times 142=2130

Answer

21302130 bricks

Q11. If S1,S2,S3,…,SmS_1, S_2, S_3, \ldots, S_m are the sums of nn terms of mm A.P.'s whose …

If S1,S2,S3,…,SmS_1, S_2, S_3, \ldots, S_m are the sums of nn terms of mm A.P.'s whose first terms are 1,2,3,…,m1, 2, 3, \ldots, m and whose common differences are 1,3,5,…,(2m−1)1, 3, 5, \ldots, (2m-1) respectively, then show that S1+S2+S3+⋯+Sm=12mn(mn+1)S_1 + S_2 + S_3 + \cdots + S_m = \dfrac{1}{2} mn(mn+1).

Show steps (8)

  1. The kthk^{th} A.P. has first term kk and common difference 2k−12k-1.

  2. Sk=n2[2k+(n−1)(2k−1)]S_k=\dfrac{n}{2}[2k+(n-1)(2k-1)]

  3. S1+S2+⋯+Sm=n2∑k=1m[2k+(n−1)(2k−1)]S_1+S_2+\cdots+S_m=\dfrac{n}{2}\sum_{k=1}^{m}\big[2k+(n-1)(2k-1)\big]

  4. =n2[2⋅m(m+1)2+(n−1)(2⋅m(m+1)2−m)]=\dfrac{n}{2}\Big[2\cdot\dfrac{m(m+1)}{2}+(n-1)\big(2\cdot\dfrac{m(m+1)}{2}-m\big)\Big]

  5. =n2[m(m+1)+(n−1)(m2+m−m)]=\dfrac{n}{2}\big[m(m+1)+(n-1)(m^2+m-m)\big]

  6. =n2[m2+m+(n−1)m2]=\dfrac{n}{2}\big[m^2+m+(n-1)m^2\big]

  7. =n2[m+nm2]=mn2(1+mn)=\dfrac{n}{2}\big[m+nm^2\big]=\dfrac{mn}{2}(1+mn)

  8. ∴S1+S2+⋯+Sm=12mn(mn+1)\therefore S_1+S_2+\cdots+S_m=\dfrac{1}{2}mn(mn+1)

Answer

S1+S2+⋯+Sm=12mn(mn+1)S_1+S_2+\cdots+S_m=\dfrac{1}{2}mn(mn+1) is proved.

Q12. Find the sum [a−ba+b+3a−2ba+b+5a−3ba+b+⋯to 12 terms]\left[ \dfrac{a-b}{a+b} + \dfrac{3a-2b}{a+b} + \dfrac{5a-3b}{a+b} + \cdots \text{to } 12 \text{ terms} \right] .

Show steps (6)

  1. Terms: a−ba+b, 3a−2ba+b, 5a−3ba+b,…\dfrac{a-b}{a+b},\ \dfrac{3a-2b}{a+b},\ \dfrac{5a-3b}{a+b},\ldots

  2. First term A=a−ba+bA=\dfrac{a-b}{a+b}

  3. Common difference D=3a−2ba+b−a−ba+b=2a−ba+bD=\dfrac{3a-2b}{a+b}-\dfrac{a-b}{a+b}=\dfrac{2a-b}{a+b}

  4. S12=122[2⋅a−ba+b+11⋅2a−ba+b]S_{12}=\dfrac{12}{2}\left[2\cdot\dfrac{a-b}{a+b}+11\cdot\dfrac{2a-b}{a+b}\right]

  5. =6[2a−2b+22a−11ba+b]=6\left[\dfrac{2a-2b+22a-11b}{a+b}\right]

  6. =6(24a−13b)a+b=144a−78ba+b=\dfrac{6(24a-13b)}{a+b}=\dfrac{144a-78b}{a+b}

Answer

6(24a−13b)a+b=144a−78ba+b\dfrac{6(24a-13b)}{a+b}=\dfrac{144a-78b}{a+b}

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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