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Class 10 · Tamil Nadu State Board

Samacheer Kalvi 10th Maths Numbers and Sequences Exercise Solutions | KalviGavani

TN Class 10 Maths Numbers and Sequences exercise solutions — Euclid division, HCF, LCM, modular arithmetic, AP nth term, AP sum, GP, special series, full step-by-step answers.

Updated 15 June 202632 min read

How to Use This Page

Solve each Samacheer Kalvi exercise question on your own first, then compare your method and answer with the step-by-step solutions below. Every question from Chapter 2 (Numbers and Sequences) is worked out in full. Show every division step in Euclid's algorithm problems, write the AP formula before substituting values, and verify your answers. Pay special attention to the Common Mistakes section to avoid losing marks.

Key Concepts (Quick Recap)

  • Euclid's Division Lemma: For any positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 <= r < b
  • Euclid's Division Algorithm: To find HCF(a, b), repeatedly apply the lemma until r = 0. The last non-zero remainder is the HCF
  • Fundamental Theorem of Arithmetic: Every composite number can be expressed as a product of primes, and this factorization is unique (apart from order)
  • HCF x LCM = a x b: For two positive integers, the product of HCF and LCM equals the product of the numbers
  • Modular Arithmetic: a is congruent to b (mod m) means m divides (a - b); both leave the same remainder when divided by m
  • Sequence: An arrangement of numbers in a definite order according to some rule
  • Arithmetic Progression (AP): A sequence where the common difference d = t(n) - t(n-1) is constant
  • nth term of AP: t_n = a + (n - 1)d
  • Sum of n terms of AP: S_n = n/2 [2a + (n - 1)d] or S_n = n/2 (a + l) where l is the last term
  • Geometric Progression (GP): A sequence where the common ratio r = t(n)/t(n-1) is constant
  • nth term of GP: t_n = ar^(n-1)
  • Sum of n terms of GP: S_n = a(r^n - 1)/(r - 1) for r > 1; S_n = a(1 - r^n)/(1 - r) for r < 1
  • Sum of infinite GP: S = a/(1 - r) for |r| < 1
  • Special Series: Sum of n natural numbers = n(n+1)/2; Sum of squares = n(n+1)(2n+1)/6; Sum of cubes = [n(n+1)/2]^2

Exercise 2.1 — Euclid's Division Lemma and HCF

Q1. Find all positive integers which, when divided by 3, leave a remainder of 2.

Solution
By Euclid's Division Lemma: a = bq + r, where 0 <= r < b Here b = 3, r = 2. a = 3q + 2, where q = 0, 1, 2, 3, ... When q = 0: a = 3(0) + 2 = 2 When q = 1: a = 3(1) + 2 = 5 When q = 2: a = 3(2) + 2 = 8 When q = 3: a = 3(3) + 2 = 11 The numbers are: 2, 5, 8, 11, 14, ... General form: a = 3q + 2, q >= 0
AnswerAll positive integers of the form 3q + 2, i.e. 2, 5, 8, 11, 14, ...

Q2. A gardener has 532 flower pots. He arranges them in rows of 21. How many complete rows can he make? How many pots are left over?

Solution
532 = 21 x q + r 532 / 21 = 25 remainder 7 532 = 21 x 25 + 7 Complete rows = 25 Pots left over = 7
Answer25 complete rows, 7 pots left over

Q3. Prove that the product of two consecutive positive integers is divisible by 2.

Solution
Let the two consecutive positive integers be n and n + 1. Case 1: n is even n = 2k for some positive integer k. Product = n(n+1) = 2k(2k+1) = 2[k(2k+1)] This is clearly divisible by 2. Case 2: n is odd n = 2k - 1 for some positive integer k. n + 1 = 2k (which is even) Product = n(n+1) = (2k-1)(2k) = 2[k(2k-1)] This is also divisible by 2. In both cases, the product is divisible by 2. Proved.
AnswerProved. The product of two consecutive positive integers is always divisible by 2.

Q4. If a, b, c divided by 13 give remainders 9, 7, and 10 respectively, show that a + b + c is divisible by 13.

Solution
a = 13p + 9 b = 13q + 7 c = 13r + 10 a + b + c = 13p + 9 + 13q + 7 + 13r + 10 = 13(p + q + r) + 26 = 13(p + q + r) + 13 x 2 = 13(p + q + r + 2) a + b + c is a multiple of 13. Divisible by 13. Proved.
AnswerProved. a + b + c = 13(p + q + r + 2), which is divisible by 13.

Q5. Prove that the square of any positive integer leaves remainder 0 or 1 when divided by 4.

Solution
By Euclid's Division Lemma, any positive integer n can be written as: n = 4q + r, where r = 0, 1, 2, or 3. Case 1: r = 0, n = 4q n^2 = 16q^2 = 4(4q^2) Remainder when divided by 4 = 0 Case 2: r = 1, n = 4q + 1 n^2 = (4q+1)^2 = 16q^2 + 8q + 1 = 4(4q^2 + 2q) + 1 Remainder when divided by 4 = 1 Case 3: r = 2, n = 4q + 2 n^2 = (4q+2)^2 = 16q^2 + 16q + 4 = 4(4q^2 + 4q + 1) Remainder when divided by 4 = 0 Case 4: r = 3, n = 4q + 3 n^2 = (4q+3)^2 = 16q^2 + 24q + 9 = 4(4q^2+6q+2) + 1 Remainder when divided by 4 = 1 In all cases, remainder is either 0 or 1. Proved.
AnswerProved. The square of any positive integer leaves remainder 0 or 1 when divided by 4.

Q6. Find the HCF using Euclid's algorithm: (i) 340 and 412 (ii) 867 and 255 (iii) 10224 and 9648 (iv) 84, 90, and 120

Solution
(i) HCF(340, 412): 412 = 340 x 1 + 72 340 = 72 x 4 + 52 72 = 52 x 1 + 20 52 = 20 x 2 + 12 20 = 12 x 1 + 8 12 = 8 x 1 + 4 8 = 4 x 2 + 0 HCF = 4 (ii) HCF(867, 255): 867 = 255 x 3 + 102 255 = 102 x 2 + 51 102 = 51 x 2 + 0 HCF = 51 (iii) HCF(10224, 9648): 10224 = 9648 x 1 + 576 9648 = 576 x 16 + 432 576 = 432 x 1 + 144 432 = 144 x 3 + 0 HCF = 144 (iv) HCF(84, 90, 120): First find HCF(84, 90): 90 = 84 x 1 + 6 84 = 6 x 14 + 0 HCF(84, 90) = 6 Now find HCF(6, 120): 120 = 6 x 20 + 0 HCF(6, 120) = 6 HCF(84, 90, 120) = 6
Answer(i) 4; (ii) 51; (iii) 144; (iv) 6

Q7. Find the largest number that divides 1230 and 1926, leaving remainder 12 in each case.

Solution
Since remainder is 12 in both cases, subtract 12 from each number. 1230 - 12 = 1218 1926 - 12 = 1914 Now find HCF(1218, 1914): 1914 = 1218 x 1 + 696 1218 = 696 x 1 + 522 696 = 522 x 1 + 174 522 = 174 x 3 + 0 HCF = 174
Answer174

Q8. If d = HCF(32, 60), find integers x and y such that d = 32x + 60y.

Solution
HCF(32, 60): 60 = 32 x 1 + 28 32 = 28 x 1 + 4 28 = 4 x 7 + 0 d = HCF = 4 Now express 4 using back-substitution: 4 = 32 - 28 x 1 = 32 - (60 - 32 x 1) x 1 = 32 - 60 + 32 = 2(32) + (-1)(60) So x = 2, y = -1. Check: 32(2) + 60(-1) = 64 - 60 = 4 = d. Correct.
Answerx = 2, y = -1

Q9. A number when divided by 88 gives remainder 61. What is the remainder when this number is divided by 11?

Solution
Let the number be N. N = 88q + 61 Now divide N by 11: N = 88q + 61 = 11 x 8q + 61 61 divided by 11: 61 = 11 x 5 + 6 Remainder = 6 (Also: Since 88 is a multiple of 11, the remainder of N divided by 11 is the same as the remainder of 61 divided by 11, which is 6.)
Answer6

Q10. Prove that two consecutive positive integers are always coprime.

Solution
Let the two consecutive integers be n and n + 1. Suppose d is a common factor of n and n + 1. Then d divides n and d divides (n + 1). So d divides (n + 1) - n = 1. The only positive integer that divides 1 is 1 itself. So d = 1, which means n and n + 1 are coprime. Proved.
AnswerProved. Two consecutive positive integers always have HCF = 1.

Exercise 2.2 — Fundamental Theorem of Arithmetic

Q1. For what values of n does 4 raised to n end with the digit 6?

Solution
4^1 = 4 (ends in 4) 4^2 = 16 (ends in 6) 4^3 = 64 (ends in 4) 4^4 = 256 (ends in 6) 4^5 = 1024 (ends in 4) Pattern: When n is odd, 4^n ends in 4. When n is even, 4^n ends in 6. 4^n ends with digit 6 when n is even (n = 2, 4, 6, ...).
Answern must be a positive even integer.

Q2. For what values of m does 2 raised to n times 5 raised to m end with the digit 5?

Solution
2^n x 5^m ends in 5 only if the units digit is 5. 5^1 = 5, 5^2 = 25, 5^3 = 125 — all end in 5. 2^n always gives an even number (ends in 2, 4, 6, or 8). For the product to end in 5, we need 2^n x 5^m. 5^m always ends in 5. But multiplying by 2^n (which is even) will make the last digit even, not 5. Exception: If n = 0, then 2^0 = 1. Then 2^0 x 5^m = 5^m, which ends in 5 for m >= 1. So n = 0 and m >= 1. But if n > 0, the product 2^n x 5^m = 2^n-m x 10^m (when n >= m) or = 5^(m-n) x 10^n (when m > n). When m > n: product = 5^(m-n) x 10^n, ends in 0 (not 5). When n >= m: product = 2^(n-m) x 10^m, ends in 0 (not 5). When n = 0: product = 5^m, ends in 5. Answer: n = 0 and m >= 1.
Answern must be 0 (i.e., 2^n factor is absent) and m >= 1.

Q3. Express 13824 as 2 raised to a times 3 raised to b, and find a and b.

Solution
13824 / 2 = 6912 6912 / 2 = 3456 3456 / 2 = 1728 1728 / 2 = 864 864 / 2 = 432 432 / 2 = 216 216 / 2 = 108 108 / 2 = 54 54 / 2 = 27 27 / 3 = 9 9 / 3 = 3 3 / 3 = 1 13824 = 2^9 x 3^3 a = 9, b = 3
Answera = 9, b = 3

Q4. Express 113400 in prime factorization form.

Solution
113400 / 2 = 56700 56700 / 2 = 28350 28350 / 2 = 14175 14175 / 3 = 4725 4725 / 3 = 1575 1575 / 3 = 525 525 / 3 = 175 175 / 5 = 35 35 / 5 = 7 7 / 7 = 1 113400 = 2^3 x 3^4 x 5^2 x 7^1
Answer113400 = 2^3 x 3^4 x 5^2 x 7^1

Q5. Find the LCM and HCF of 408 and 170.

Solution
HCF by Euclid's algorithm: 408 = 170 x 2 + 68 170 = 68 x 2 + 34 68 = 34 x 2 + 0 HCF = 34 LCM = (408 x 170) / HCF = 69360 / 34 = 2040
AnswerHCF = 34, LCM = 2040

Q6. Find the greatest 6-digit number divisible by 24, 15, and 36.

Solution
First find LCM(24, 15, 36): 24 = 2^3 x 3 15 = 3 x 5 36 = 2^2 x 3^2 LCM = 2^3 x 3^2 x 5 = 8 x 9 x 5 = 360 Greatest 6-digit number = 999999 999999 / 360 = 2777 remainder 279 999999 - 279 = 999720 Check: 999720 / 24 = 41655, 999720 / 15 = 66648, 999720 / 36 = 27770 All exact.
Answer999720

Q7. Find the smallest number that leaves remainder 7 when divided by 35, 56, and 91.

Solution
The required number = LCM(35, 56, 91) + 7 35 = 5 x 7 56 = 2^3 x 7 91 = 7 x 13 LCM = 2^3 x 5 x 7 x 13 = 8 x 5 x 7 x 13 = 3640 Number = 3640 + 7 = 3647
Answer3647

Q8. Find the least number divisible by the first 10 natural numbers.

Solution
First 10 natural numbers: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 LCM = LCM(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) Prime factorization: 1 = 1 2 = 2 3 = 3 4 = 2^2 5 = 5 6 = 2 x 3 7 = 7 8 = 2^3 9 = 3^2 10 = 2 x 5 Take highest powers: 2^3, 3^2, 5, 7 LCM = 8 x 9 x 5 x 7 = 2520
Answer2520

Exercise 2.3 — Modular Arithmetic

Q1. Solve: (i) 71 is congruent to x (mod 8) (ii) 78 + x is congruent to 3 (mod 5) (iii) 89 is congruent to (x + 3) (mod 4) (iv) 96 is congruent to x/7 (mod 5) (v) 5x is congruent to 4 (mod 6)

Solution
(i) 71 mod 8: 71 / 8 = 8 remainder 7 71 = 8 x 8 + 7 x = 7 (ii) 78 + x = 3 (mod 5) 78 mod 5 = 3 (since 78 = 15 x 5 + 3) So 3 + x = 3 (mod 5) x = 0 (mod 5) Smallest positive x = 5 (iii) 89 = (x + 3) (mod 4) 89 mod 4 = 1 (since 89 = 22 x 4 + 1) x + 3 = 1 (mod 4) x = -2 (mod 4) x = 2 (mod 4) Smallest positive x = 2 (iv) 96 = x/7 (mod 5) 96 mod 5 = 1 (since 96 = 19 x 5 + 1) x/7 = 1 (mod 5) x = 7 (mod 5) x mod 5 = 2 But x/7 must be a valid integer. x = 7 gives x/7 = 1, and 1 mod 5 = 1. Correct. x = 7 (v) 5x = 4 (mod 6) Try x = 1: 5(1) = 5, 5 mod 6 = 5 (no) Try x = 2: 5(2) = 10, 10 mod 6 = 4 (yes!) x = 2 (mod 6) Solutions: x = 2, 8, 14, 20, ...
Answer(i) x = 7; (ii) x = 5; (iii) x = 2; (iv) x = 7; (v) x = 2, 8, 14, 20, ...

Q2. What will be the time 100 hours after 7 AM?

Solution
100 hours = 4 days + 4 hours (since 100 = 24 x 4 + 4) 7 AM + 4 hours = 11 AM The time after 100 hours is 11 AM.
Answer11 AM

Q3. What day will it be 45 days after Tuesday?

Solution
45 = 7 x 6 + 3 45 days = 6 complete weeks + 3 extra days. 6 weeks after Tuesday is still Tuesday. 3 days after Tuesday = Friday. The day will be Friday.
AnswerFriday

Q4. Find the remainder when 2 raised to 81 is divided by 17.

Solution
2^81 mod 17 By Fermat's Little Theorem: 2^16 = 1 (mod 17) (Since 17 is prime and 2 is not divisible by 17) 81 = 16 x 5 + 1 2^81 = 2^(16x5+1) = (2^16)^5 x 2^1 = 1^5 x 2 = 2 (mod 17) Remainder = 2
Answer2

Q5. Prove that 2 raised to n plus 6 times 9 raised to n is divisible by 7 for all natural numbers n.

Solution
2^n + 6(9^n) = 2^n + 6(9^n) 9 mod 7 = 2, so 9^n mod 7 = 2^n mod 7 2^n + 6(9^n) mod 7 = 2^n + 6(2^n) mod 7 = 2^n(1 + 6) mod 7 = 2^n x 7 mod 7 = 0 mod 7 So 2^n + 6(9^n) is always divisible by 7. Proved.
AnswerProved. 2^n + 6(9^n) = 7(2^n), which is divisible by 7.

Exercise 2.4 — Sequences

Q1. Find the next three terms of the following sequences: (i) 8, 24, 72, ... (ii) 5, 1, -3, ... (iii) 1/4, 2/9, 3/16, ...

Solution
(i) 8, 24, 72, ... 24/8 = 3, 72/24 = 3 Common ratio = 3 (geometric sequence) Next terms: 72 x 3 = 216, 216 x 3 = 648, 648 x 3 = 1944 Next three terms: 216, 648, 1944 (ii) 5, 1, -3, ... 1 - 5 = -4, -3 - 1 = -4 Common difference = -4 (arithmetic sequence) Next terms: -3 + (-4) = -7, -7 + (-4) = -11, -11 + (-4) = -15 Next three terms: -7, -11, -15 (iii) 1/4, 2/9, 3/16, ... Rewrite: 1/(2^2), 2/(3^2), 3/(4^2), ... nth term = n/(n+1)^2 Next: 4/(5^2) = 4/25, 5/(6^2) = 5/36, 6/(7^2) = 6/49 Next three terms: 4/25, 5/36, 6/49
Answer(i) 216, 648, 1944; (ii) -7, -11, -15; (iii) 4/25, 5/36, 6/49

Q2. Find the nth term of the following sequences: (i) 2, 5, 10, 17, ... (ii) 0, 1/2, 2/3, 3/4, ... (iii) 3, 8, 13, 18, ...

Solution
(i) 2, 5, 10, 17, ... Differences: 3, 5, 7, ... (odd numbers starting from 3) Try n^2: 1, 4, 9, 16, ... Each term = n^2 + 1 Check: 1^2+1=2, 2^2+1=5, 3^2+1=10, 4^2+1=17. Correct. nth term = n^2 + 1 (ii) 0, 1/2, 2/3, 3/4, ... Write as: 0/1, 1/2, 2/3, 3/4, ... Pattern: numerator = n - 1, denominator = n nth term = (n - 1)/n (iii) 3, 8, 13, 18, ... Differences: 5, 5, 5, ... (AP with d = 5) This is an AP with a = 3, d = 5 nth term = a + (n-1)d = 3 + 5(n-1) = 3 + 5n - 5 = 5n - 2 Check: 5(1)-2=3, 5(2)-2=8, 5(3)-2=13. Correct.
Answer(i) n^2 + 1; (ii) (n-1)/n; (iii) 5n - 2

Q3. If a1 = 1, a2 = 1, and a_n = 2a_(n-1) + a_(n-2), find the first 6 terms.

Solution
a1 = 1 a2 = 1 a3 = 2(a2) + a1 = 2(1) + 1 = 3 a4 = 2(a3) + a2 = 2(3) + 1 = 7 a5 = 2(a4) + a3 = 2(7) + 3 = 17 a6 = 2(a5) + a4 = 2(17) + 7 = 41 First 6 terms: 1, 1, 3, 7, 17, 41
Answer1, 1, 3, 7, 17, 41

Exercise 2.5 — Arithmetic Progression

Q1. Check whether the following sequences are AP: (i) a - 3, a - 5, a - 7 (ii) 1/2, 1/3, 1/4 (iii) 9, 13, 17, 21, 25 (iv) -1/3, 0, 1/3, 2/3 (v) 1, -1, 1, -1

Solution
(i) a-3, a-5, a-7 d1 = (a-5)-(a-3) = -2 d2 = (a-7)-(a-5) = -2 d1 = d2 = -2. YES, it is an AP. (ii) 1/2, 1/3, 1/4 d1 = 1/3 - 1/2 = (2-3)/6 = -1/6 d2 = 1/4 - 1/3 = (3-4)/12 = -1/12 d1 is not equal to d2. NOT an AP. (iii) 9, 13, 17, 21, 25 d1 = 13-9 = 4, d2 = 17-13 = 4, d3 = 21-17 = 4, d4 = 25-21 = 4 All equal. YES, it is an AP with d = 4. (iv) -1/3, 0, 1/3, 2/3 d1 = 0-(-1/3) = 1/3, d2 = 1/3-0 = 1/3, d3 = 2/3-1/3 = 1/3 All equal. YES, it is an AP with d = 1/3. (v) 1, -1, 1, -1 d1 = -1-1 = -2, d2 = 1-(-1) = 2 d1 is not equal to d2. NOT an AP.
Answer(i) Yes (d=-2); (ii) No; (iii) Yes (d=4); (iv) Yes (d=1/3); (v) No

Q2. Find the 19th term of the AP: -11, -15, -19, ...

Solution
a = -11, d = -15-(-11) = -4, n = 19 t_n = a + (n - 1)d t_19 = -11 + (19 - 1)(-4) = -11 + 18(-4) = -11 - 72 = -83
Answer-83

Q3. Which term of the AP 16, 11, 6, 1, ... is -54?

Solution
a = 16, d = 11 - 16 = -5, t_n = -54 t_n = a + (n-1)d -54 = 16 + (n-1)(-5) -54 = 16 - 5n + 5 -54 = 21 - 5n 5n = 21 + 54 = 75 n = 15 The 15th term is -54.
Answer15th term

Q4. Find the middle terms of the AP: 9, 15, 21, ..., 183.

Solution
a = 9, d = 6, last term l = 183 l = a + (n-1)d 183 = 9 + (n-1)(6) 174 = (n-1)(6) n - 1 = 29 n = 30 Since n = 30 (even), there are two middle terms: n/2 = 15th term and (n/2 + 1) = 16th term t_15 = 9 + (15-1)(6) = 9 + 84 = 93 t_16 = 9 + (16-1)(6) = 9 + 90 = 99 Middle terms: 93 and 99
Answer93 and 99

Q5. If 3 + k, 18 - k, 5k + 1 are in AP, find k.

Solution
If three terms are in AP, then: 2(middle term) = first term + third term 2(18 - k) = (3 + k) + (5k + 1) 36 - 2k = 3 + k + 5k + 1 36 - 2k = 6k + 4 32 = 8k k = 4 Check: terms are 7, 14, 21. Common difference = 7. Yes, AP.
Answerk = 4

Q6. Three consecutive terms of an AP have sum 27 and product 288. Find the terms.

Solution
Let the three consecutive terms be (a - d), a, (a + d). Sum: (a - d) + a + (a + d) = 3a = 27 a = 9 Product: (9 - d)(9)(9 + d) = 288 9(81 - d^2) = 288 81 - d^2 = 32 d^2 = 49 d = 7 Terms: 9 - 7 = 2, 9, 9 + 7 = 16 The three terms are 2, 9, 16. Check: Sum = 2 + 9 + 16 = 27. Product = 2 x 9 x 16 = 288. Correct.
Answer2, 9, 16

Q7. If the ratio of the 6th term to the 8th term of an AP is 7:9, find the ratio of the 9th term to the 13th term.

Solution
t_6 = a + 5d, t_8 = a + 7d t_6/t_8 = (a + 5d)/(a + 7d) = 7/9 9(a + 5d) = 7(a + 7d) 9a + 45d = 7a + 49d 2a = 4d a = 2d t_9 = a + 8d = 2d + 8d = 10d t_13 = a + 12d = 2d + 12d = 14d t_9/t_13 = 10d/14d = 5/7
Answer5:7

Q8. An auditorium has 30 seats in the first row, 32 in the second, 34 in the third, and so on for 30 rows. How many seats are in the last row?

Solution
a = 30, d = 2, n = 30 t_30 = a + (n-1)d = 30 + (30-1)(2) = 30 + 58 = 88
Answer88 seats

Exercise 2.6 — Sum of AP

Q1. Find the sum of: (i) 3 + 7 + 11 + ... to 40 terms (ii) 102 + 97 + 92 + ... to 27 terms

Solution
(i) a = 3, d = 4, n = 40 S_n = n/2 [2a + (n-1)d] S_40 = 40/2 [2(3) + (40-1)(4)] = 20 [6 + 39(4)] = 20 [6 + 156] = 20 x 162 = 3240 (ii) a = 102, d = -5, n = 27 S_27 = 27/2 [2(102) + (27-1)(-5)] = 27/2 [204 + 26(-5)] = 27/2 [204 - 130] = 27/2 x 74 = 27 x 37 = 999
Answer(i) 3240; (ii) 999

Q2. How many consecutive odd integers starting from 5 must be added to get 480?

Solution
Odd integers starting from 5: 5, 7, 9, 11, ... This is an AP with a = 5, d = 2, S_n = 480 S_n = n/2 [2a + (n-1)d] 480 = n/2 [2(5) + (n-1)(2)] 480 = n/2 [10 + 2n - 2] 480 = n/2 [2n + 8] 480 = n(n + 4) n^2 + 4n - 480 = 0 n = [-4 + sqrt(16 + 1920)] / 2 = [-4 + sqrt(1936)] / 2 = [-4 + 44] / 2 = 20 20 consecutive odd integers must be added.
Answer20

Q3. Find the sum of the first 28 terms of an AP whose nth term is t_n = 4n - 3.

Solution
t_n = 4n - 3 First term: t_1 = 4(1) - 3 = 1 28th term: t_28 = 4(28) - 3 = 109 S_n = n/2 (first term + last term) S_28 = 28/2 (1 + 109) = 14 x 110 = 1540
Answer1540

Q4. Find the sum of all odd positive integers less than 450.

Solution
Odd positive integers less than 450: 1, 3, 5, ..., 449 This is an AP with a = 1, d = 2, last term l = 449 l = a + (n-1)d 449 = 1 + (n-1)(2) 448 = 2(n-1) n = 225 S_n = n/2 (a + l) = 225/2 (1 + 449) = 225/2 x 450 = 225 x 225 = 50625
Answer50625

Q5. A man repays a loan of Rs 65,000 by paying Rs 500 in the first month and then increasing the payment by Rs 100 each month. How long will it take to repay the loan?

Solution
Monthly payments: 500, 600, 700, 800, ... This is an AP with a = 500, d = 100 Total to repay = 65000 S_n = n/2 [2a + (n-1)d] 65000 = n/2 [2(500) + (n-1)(100)] 65000 = n/2 [1000 + 100n - 100] 65000 = n/2 [100n + 900] 65000 = n(50n + 450) 1300 = 5n^2 + 45n 5n^2 + 45n - 1300 = 0 n^2 + 9n - 260 = 0 (n + 20)(n - 13) = 0 n = 13 (n cannot be negative) It will take 13 months.
Answer13 months

Exercise 2.7 — Geometric Progression

Q1. Identify which of the following are GP: (i) 2, 6, 18, 54 (ii) 1, -1, 1, -1 (iii) 1, 4, 9, 16 (iv) 5, 10, 15, 20

Solution
(i) 2, 6, 18, 54 6/2 = 3, 18/6 = 3, 54/18 = 3 Common ratio r = 3. YES, GP. (ii) 1, -1, 1, -1 -1/1 = -1, 1/(-1) = -1, (-1)/1 = -1 Common ratio r = -1. YES, GP. (iii) 1, 4, 9, 16 4/1 = 4, 9/4 = 2.25 Ratios not equal. NOT a GP. (iv) 5, 10, 15, 20 10/5 = 2, 15/10 = 1.5 Ratios not equal. NOT a GP. (This is an AP with d = 5)
Answer(i) Yes (r=3); (ii) Yes (r=-1); (iii) No; (iv) No

Q2. Find the 7th term of the GP: 729, 243, 81, ...

Solution
a = 729, r = 243/729 = 1/3 t_n = ar^(n-1) t_7 = 729 x (1/3)^6 = 729 x (1/729) = 1
Answer1

Q3. If the 9th term of a GP is 512 and the 6th term is 64, find the first term and common ratio.

Solution
t_9 = ar^8 = 512 ...(1) t_6 = ar^5 = 64 ...(2) Dividing (1) by (2): r^3 = 512/64 = 8 r = cube root of 8 = 2 From (2): a(2)^5 = 64 32a = 64 a = 2
Answera = 2, r = 2

Q4. Three consecutive terms of a GP have product 27 and sum of pairwise products 57/2. Find the terms.

Solution
Let the terms be a/r, a, ar. Product: (a/r)(a)(ar) = a^3 = 27 a = 3 Sum of pairwise products: (a/r)(a) + (a)(ar) + (a/r)(ar) = a^2/r + a^2 r + a^2 = a^2(1/r + r + 1) = 9(1/r + r + 1) = 57/2 1/r + r + 1 = 57/18 = 19/6 Multiply by r: 1 + r^2 + r = 19r/6 6 + 6r^2 + 6r = 19r 6r^2 - 13r + 6 = 0 r = [13 + sqrt(169-144)] / 12 = [13 + 5]/12 or [13-5]/12 r = 18/12 = 3/2 or r = 8/12 = 2/3 When r = 3/2: terms are 2, 3, 9/2 When r = 2/3: terms are 9/2, 3, 2 The three terms are 2, 3, 9/2 (or equivalently 9/2, 3, 2).
Answer2, 3, 9/2

Exercise 2.8 — Sum of GP

Q1. Find the sum of the first 6 terms of the GP: 5, 15, 45, ...

Solution
a = 5, r = 15/5 = 3, n = 6 S_n = a(r^n - 1)/(r - 1) S_6 = 5(3^6 - 1)/(3 - 1) = 5(729 - 1)/2 = 5 x 728/2 = 5 x 364 = 1820
Answer1820

Q2. Find the sum to infinity of the GP: 1, 1/3, 1/9, 1/27, ...

Solution
a = 1, r = 1/3, |r| < 1 S = a/(1 - r) = 1/(1 - 1/3) = 1/(2/3) = 3/2
Answer3/2

Q3. Find the sum of the GP: 3 + 6 + 12 + ... + 1536.

Solution
a = 3, r = 2, last term l = 1536 l = ar^(n-1) 1536 = 3(2)^(n-1) 512 = 2^(n-1) 2^9 = 2^(n-1) n - 1 = 9, n = 10 S_10 = 3(2^10 - 1)/(2 - 1) = 3(1024 - 1) = 3 x 1023 = 3069
Answer3069

Q4. Express 0.bar(123) as a rational number.

Solution
x = 0.123123123... Multiply by 1000: 1000x = 123.123123... Subtract: 1000x - x = 123 999x = 123 x = 123/999 = 41/333
Answer41/333

Exercise 2.9 — Special Series

Q1. Find the sum of the first 60 natural numbers.

Solution
S = n(n+1)/2 = 60(61)/2 = 1830
Answer1830

Q2. Find the sum of the squares of the first 15 natural numbers.

Solution
S = n(n+1)(2n+1)/6 = 15(16)(31)/6 = 7440/6 = 1240
Answer1240

Q3. Find the sum of the cubes of the first 10 natural numbers.

Solution
S = [n(n+1)/2]^2 = [10(11)/2]^2 = [55]^2 = 3025
Answer3025

Q4. If the sum of the first n natural numbers is 325, find n.

Solution
n(n+1)/2 = 325 n(n+1) = 650 n^2 + n - 650 = 0 n = [-1 + sqrt(1 + 2600)]/2 = [-1 + sqrt(2601)]/2 = [-1 + 51]/2 = 25
Answern = 25

Q5. Find the sum of (2^3 - 1^3) + (4^3 - 3^3) + (6^3 - 5^3) + ... to 8 terms.

Solution
Each term is (2k)^3 - (2k-1)^3 for k = 1, 2, 3, ... (2k)^3 - (2k-1)^3 = 8k^3 - (8k^3 - 12k^2 + 6k - 1) = 12k^2 - 6k + 1 Sum of 8 terms = sum of (12k^2 - 6k + 1) for k = 1 to 8 = 12 x sum(k^2) - 6 x sum(k) + sum(1) for k = 1 to 8 sum(k) = 8(9)/2 = 36 sum(k^2) = 8(9)(17)/6 = 204 = 12(204) - 6(36) + 8 = 2448 - 216 + 8 = 2240
Answer2240

Q6. Find the sum of the odd numbers: 1 + 3 + 5 + ... + 71.

Solution
Odd numbers from 1 to 71 form an AP: a = 1, d = 2, l = 71 l = a + (n-1)d 71 = 1 + (n-1)(2) 70 = 2(n-1) n = 36 S_n = n/2(a + l) = 36/2(1 + 71) = 18 x 72 = 1296 (Alternatively: Sum of first n odd numbers = n^2. Here n = 36, so 36^2 = 1296.)
Answer1296

Q7. If the sum of the first n natural numbers is 325, find the sum of the cubes of the first n natural numbers.

Solution
n(n+1)/2 = 325 n = 25 (from Q4 above) Sum of cubes = [n(n+1)/2]^2 = [325]^2 = 105625
Answer105625

Common Mistakes

1. Stopping Euclid's algorithm before remainder becomes zero. The algorithm must continue until the remainder is exactly 0. The HCF is the LAST non-zero remainder. If you stop one step early, you get the wrong answer.

2. Using n instead of (n-1) in the AP nth term formula. The formula is t_n = a + (n-1)d, NOT a + nd. For example, the 5th term of AP 2, 5, 8, ... is a + 4d = 2 + 12 = 14, not a + 5d = 17.

3. Forgetting the sign of a negative common difference. In AP 24, 21, 18, ..., d = -3, not 3. Using d = 3 will give completely wrong answers for every subsequent calculation.

4. Confusing nth term and sum formula. Use t_n = a + (n-1)d for finding a specific term. Use S_n = n/2[2a + (n-1)d] for finding the total. Using the wrong formula is an instant mark-loser.

5. Computing LCM incorrectly by multiplying all prime factors. LCM takes the HIGHEST power of each prime, not the sum or product of all factors. LCM of 24 and 36: 24 = 2^3 x 3, 36 = 2^2 x 3^2. LCM = 2^3 x 3^2 = 72, not 2^(3+2) x 3^(1+2).

6. Dividing by r in GP sum formula when r = 1. If r = 1, every term equals a, so S_n = na. The formula a(r^n - 1)/(r-1) gives 0/0 when r = 1, which is undefined.

7. Using the wrong sum formula for infinite GP. S = a/(1-r) only works whenr< 1. Ifr>= 1, the infinite sum does not exist. Students sometimes apply this formula regardless.

8. Not verifying answers by substitution. After finding the 15th term, plug it back into the formula to check. After finding HCF, verify it divides both original numbers. Simple verification catches arithmetic errors.

Exam Checklist

  • [ ] I can apply Euclid's Division Lemma to express a = bq + r.
  • [ ] I can find HCF using Euclid's division algorithm step by step.
  • [ ] I can prove divisibility properties using the lemma.
  • [ ] I can express numbers as products of primes (Fundamental Theorem of Arithmetic).
  • [ ] I can find HCF and LCM using prime factorization and the relation HCF x LCM = a x b.
  • [ ] I can solve modular arithmetic problems (congruences).
  • [ ] I can find the day or time after a given number of hours/days using modulo.
  • [ ] I can identify sequences and find next terms.
  • [ ] I can find the nth term formula for a given sequence.
  • [ ] I can check if a sequence is an AP by verifying common difference.
  • [ ] I can find the nth term of an AP using t_n = a + (n-1)d.
  • [ ] I can find the sum of n terms of an AP using S_n = n/2[2a + (n-1)d] or S_n = n/2(a+l).
  • [ ] I can solve AP word problems (instalments, seating, savings).
  • [ ] I can identify a GP and find the common ratio.
  • [ ] I can find the nth term of a GP using t_n = ar^(n-1).
  • [ ] I can find the sum of n terms and infinite sum of a GP.
  • [ ] I can apply special series formulas: sum of n, sum of n^2, sum of n^3.
  • [ ] I can convert repeating decimals to fractions using the GP method.

Internal Links

  • [Class 10 Maths — Numbers and Sequences Chapter Guide](/school/tamil-nadu-state-board/class/10/maths/numbers-and-sequences)
  • [All Class 10 Maths Solutions](/school/tamil-nadu-state-board/class/10/maths/solutions)
  • [Class 10 Maths Subject Guide](/school/tamil-nadu-state-board/class/10/maths)

Frequently Asked Questions

Apply a = bq + r repeatedly. Replace the dividend with divisor and divisor with remainder. Continue until remainder = 0. The last non-zero remainder is the HCF. Example: HCF(340,412): 412=340x1+72, 340=72x4+52, 72=52x1+20, 52=20x2+12, 20=12x1+8, 12=8x1+4, 8=4x2+0. HCF=4.
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Chapter Notes

Tamil Nadu Class 10 Maths Chapter 2 - Numbers and Sequences

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