**Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, unit, exercise 2,**

Question 1.

Prove that n^{2} – n divisible by 2 for every positive integer n.

Answer:

We know that any positive integer is of the form 2q or 2q + 1 for some integer q.

Case 1: When n = 2 q

n^{2} – n = (2q)^{2} – 2q = 4q^{2} – 2q

= 2q (2q – 1)

In n^{2} – n = 2r

2r = 2q(2q – 1)

r = q(2q + 1)

n^{2} – n is divisible by 2

Case 2: When n = 2q + 1

n^{2} – n = (2q + 1)^{2} – (2q + 1)

= 4q^{2} + 1 + 4q – 2q – 1 = 4q^{2} + 2q

= 2q (2q + 1)

If n^{2} – n = 2r

r = q (2q + 1)

∴ n^{2} – n is divisible by 2 for every positive integer “n”

Question 2.

A milk man has 175 litres of cow’s milk and 105 litres of buffalow’s milk. He wishes to sell the milk by filling the two types of milk in cans of equal capacity. Calculate the following

(i) Capacity of a can

(ii) Number of cans of cow’s milk

(iii) Number of cans of buffalow’s milk.

Answer:

175 litres of cow’s milk.

105 litres of goat’s milk.

H.C.F of 175 & 105 by using Euclid’s division algorithm.

175 = 105 × 1 + 70, the remainder 70 ≠ 0

Again using division algorithm,

105 = 70 × 1 + 35, the remainder 35 ≠ 0

Again using division algorithm.

70 = 35 × 2 + 0, the remainder is 0.

∴ 35 is the H.C.F of 175 & 105.

(i) ∴ The milk man’s milk can’s capacity is 35 litres.

(ii) No. of cow’s milk obtained = 17535 = 5 cans

(iii) No. of buffalow’s milk obtained = 10535= 3 cans

Question 3.

When the positive integers a, b and c are divided by 13 the respective remainders are 9,7 and 10. Find the remainder when a + 2b + 3c is divided by 13.

Answer:

Given the positive integer are a, b and c

a = 13q + 9 (divided by 13 leaves remainder 9)

b = 13q + 7

c = 13q + 10

a + 2b + 3c = 13q + 9 + 2(13q + 7) + 3 (13q + 10)

= 13q + 9 + 26q + 14 + 39q + 30

= 78q + 53

When compare with a = 3q + r

= (13 × 6) q + 53

The remainder is 53

Question 4.

Show that 107 is of the form 4q +3 for any integer q.

Solution:

107 = 4 × 26 + 3. This is of the form a = bq + r.

Hence it is proved.

Question 5.

If (m + 1)^{th} term of an A.P. is twice the (n + 1)^{th} term, then prove that (3m + 1)^{th} term is twice the (m + n + 1)^{th} term.

Answer:

t_{n} = a + (n – 1)d

Given t_{m+1} = 2 t_{n+1}

a + (m + 1 – 1)d = 2[a + (n + 1 – 1)d]

a + md = 2(a + nd) ⇒ a + md =2a + 2nd

md – 2nd = a

d(m – 2n) = a ….(1)

To Prove t_{(3m + 1)} = 2(t_{m+n+1})

L.H.S. = t_{3m+1}

= a + (3m + 1 – 1)d

= a + 3md

= d(m – 2n) + 3md (from 1)

= md – 2nd + 3md

= 4md – 2nd

= 2d (2m – n)

R.H.S. = 2(t_{m+n+1})

= 2 [a + (m + n + 1 – 1) d]

= 2 [a + (m + n)d]

= 2 [d (m – 2n) + md + nd)] (from 1)

= 2 [dm – 2nd + md + nd]

= 2 [2 md – nd]

= 2d (2m – n)

R.H.S = L.H.S

∴ t_{(3m+1)} = 2 t_{(m+n+1)}

Hence it is proved.

Question 6.

Find the 12^{th} term from the last term of the A.P -2, -4, -6,… -100.

Answer:

The given A.P is -2, -4, -6, …. 100

d = -4 – (-2) = -4 + 2 = – 2

Finding the 12 term from the last term

a = -100, d = 2 (taking from the last term)

n = 12

t_{n} = a + (n – 1)d

t_{12} = – 100 + 11 (2)

= -100 + 22

= -78

∴ The 12^{th} term of the A.P from the last term is – 78

Question 7.

Two A.P’s have the same common difference. The first term of one A.P is 2 and that of the other is 7. Show that the difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms.

Solution:

Let the two A.Ps be

AP_{1} = a_{1}, a_{1} + d, a_{1} + 2d,…

AP_{2} = a_{2}, a_{2} + d, a_{2} + 2d,…

In AP_{1} we have a_{1} = 2

In AP_{2} we have a_{2} = 7

t_{10} in AP_{1} = a_{1} + 9d = 2 + 9d ………….. (1)

t_{10} in AP_{2} = a_{2} + 9d = 7 + 9d …………… (2)

The difference between their 10th terms

= (1) – (2) = 2 + 9d – 7 – 9d

= -5 ………….. (I)

t_{21} m AP_{1} = a_{1} + 20d = 2 + 20d …………. (3)

t_{21} in AP_{2} = a_{2} + 20d = 7 + 20d ………… (4)

The difference between their 21 st terms is

(3) – (4)

= 2 + 20d – 7 – 20d

= -5 ……………. (II)

I = II

Hence it is Proved.

Question 8.

A man saved ₹16500 in ten years. In each year after the first he saved ₹100 more than he did in the preceding year. How much did he save in the first year?

Answer:

Amount of saving in ten years = ₹ 16500

S_{10} = 16500, d= 100

S_{n} = n2 [2a + (n – 1)d]

S_{10} = 102 [2a + 9d]

16500 = 102 [2a + 900] = 5(2a + 900)

16500 = 10a + 4500 ⇒ 16500 – 4500 = 10a

12000 = 10a

a = 1200010 = 1200

Amount saved in the first year = ₹ 1200

Question 9.

Find the G.P. in which the 2nd term is 6–√and the 6th term is 9 6–√.

Answer:

2^{nd} term of the G.P = 6–√

t_{2} = 6–√

[t_{n} = a r^{n-1}]

a.r = 6–√ ….(1)

6^{th} term of the G.P. = 9 6–√

a. r^{5} = 96–√ ……..(2)

Question 10.

The value of a motorcycle depreciates at a rate of 15% per year. What will be the value of the motorcycle 3 year hence, which is now purchased for ₹45,000?

Solution:

a = ₹45000

Depreciation = 15% for ₹45000

= 45000 × 15100

d = ₹6750 since it is depreciation

d = -6750

At the end of 1^{st} year its value = ₹45000 – ₹6750

= ₹38250,

Again depreciation = 38250 × 15100 = 5737.50

At the end of 2^{nd} year its value

= ₹38250 – ₹5737.50 = 32512.50

Again depreciation = 32512.50 × 15100 = 4876.88

At the end of the 3^{rd} year its value

= 32512.50 – 4876.88 = 27635.63

∴ The value of the automobile at the 3^{rd}year

= ₹ 27636

**Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.10,**

Question 1.

Euclid’s division lemma states that for positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy ………………….

(1) 1 < r < b

(2) 0 < r < b

(3) 0 < r < 6

(4) 0 < r < b

Ans.

(3) 0 < r < b

Question 2.

Using Euclid’s division lemma, if the cube of any positive integer is divided by 9 then the possible remainders are ………………….

(1) 0, 1, 8

(2) 1, 4, 8

(3) 0, 1, 3

(4) 1, 3, 5

Answer:

(1) 0, 1, 8

Hint: Let the +ve integer be 1, 2, 3, 4 …………

1^{3} = 1 when it is divided by 9 the remainder is 1.

2^{3} = 8 when it is divided by 9 the remainder is 8.

3^{3} = 27 when it is divided by 9 the remainder is 0.

4^{3} = 64 when it is divided by 9 the remainder is 1.

5^{3} = 125 when it is divided by 9 the remainder is 8.

The remainder 0, 1, 8 is repeated.

Question 3.

If the H.C.F of 65 and 117 is expressible in the form of 65m – 117 , then the value of m is

(1) 4

(2) 2

(3) 1

(4) 3

Answer:

(2) 2

Hint:

H.C.F. of 65 and 117

117 = 65 × 1 + 52

65 = 52 × 1 + 13

52 = 13 × 4 + 0

∴ 13 is the H.C.F. of 65 and 117.

65m – 117 = 65 × 2 – 117

130 – 117 = 13

∴ m = 2

Question 4.

The sum of the exponents of the prime factors in the prime factorization of 1729 is …………………….

(1) 1

(2) 2

(3) 3

(4) 4

Answer:

(3) 3

Hint: 1729 = 7 × 13 × 19

Sum of the exponents = 1 + 1 + 1

= 3

Question 5.

The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is

(1) 2025

(2) 5220

(3) 5025

(4) 2520

Answer:

(4) 2520

Hint:

L.C.M. = 2^{3} × 3^{2} × 5 × 7

= 8 × 9 × 5 × 7

= 2520

Question 6.

7^{4k} ≡ ______ (mod 100)

(1) 1

(2) 2

(3) 3

(4) 4

Answer:

(1) 1

Hint:

7^{4k} ≡______ (mod 100)

y^{4k} ≡ y^{4 × 1} = 1 (mod 100)

Question 7.

Given F_{1} = 1 , F_{2} = 3 and F_{n} = F_{n-1} + F_{n-2}then F_{5} is ………….

(1) 3

(2) 5

(3) 8

(4) 11

Answer:

(4) 11

Hint:

F_{n} = F_{n-1} + F_{n-2}

F_{3} = F_{2} + F_{1} = 3 + 1 = 4

F_{4} = F_{3} + F_{2} = 4 + 3 = 7

F_{5} = F_{4} + F_{3} = 7 + 4 = 11

Question 8.

The first term of an arithmetic progression is unity and the common difference is 4. Which of the following will be a term of this A.P

(1) 4551

(2) 10091

(3) 7881

(4) 13531

Answer:

(3) 7881

Hint:

t_{1} = 1

d = 4

t_{n} = a + (n – 1)d

= 1 + 4n – 4

4n – 3 = 4551

4n = 4554

n = will be a fraction

It is not possible.

4n – 3 = 10091

4n = 10091 + 3 = 10094

n = a fraction

4n – 3 = 7881

4n = 7881 + 3 = 7884

n = 78844, n is a whole number.

4n – 3 = 13531

4n = 13531 – 3 = 13534

n is a fraction.

∴ 7881 will be 1971st term of A.P.

Question 9.

If 6 times of 6^{th} term of an A.P is equal to 7 times the 7^{th} term, then the 13^{th} term of the A.P. is ………..

(1) 0

(2) 6

(3) 7

(4) 13

Answer:

(1) 0

Hint:

6 t_{6} = 7 t_{7}

6(a + 5d) = 7 (a + 6d) ⇒ 6a + 30d = 7a + 42d

30 d – 42 d = 7a – 6a ⇒ -12d = a

t_{13} = a + 12d (12d = -a)

= a – a = 0

Question 10.

An A.P consists of 31 terms. If its 16th term is m, then the sum of all the terms of this A.P. is

(1) 16 m

(2) 62 m

(3) 31 m

(4) 312 m

Answer:

(3) 31 m

Hint:

t_{16} = m

S_{31} = 312 (2a + 30d)

= 312 (2(a + 15d))

(∵ t_{16} = a + 15d)

= 31(t_{16}) = 31m

Question 11.

In an A.P., the first term is 1 and the common difference is 4. How many terms of the A.P must be taken for their sum to be equal to 120?

(1) 6

(2) 7

(3) 8

(4) 9

Answer:

(3) 8

Here a = 1, d = 4, S_{n} = 120

S_{n} = n2[2a + (n – 1)d]

120 = n2 [2 + (n – 1)4] = n2 [2 + 4n – 4)]

= n2 [4n – 2)] = n2 × 2 (2n – 1)

120 = 2n^{2} – n

∴ 2n^{2} – n – 120 = 0 ⇒ 2n^{2} – 16n + 15n – 120 = 0

2n(n – 8) + 15 (n – 8) = 0 ⇒ (n – 8) (2n + 15) = 0

n = 8 or n = −152 (omitted)

∴ n = 8

Question 12.

A = 2^{65} and B = 2^{64} + 2^{63} + 2^{62} …. + 2^{0} which of the following is true?

(1) B is 2^{64} more than A

(2) A and B are equal

(3) B is larger than A by 1

(4) A is larger than B by 1

Answer:

(4) A is larger than B by

A = 2^{65}

B = 2^{64+63} + 2^{62} + …….. + 2^{0}

= 2

= 1 + 2^{2} + 2^{2} + ……. + 2^{64}

a = 1, r = 2, n = 65 it is in G.P.

S_{65} = 1 (2^{65} – 1) = 2^{65} – 1

A = 2^{65} is larger than B

Question 13.

The next term of the sequence 316,18,112,118 is ………..

(1) 124

(2) 127

(3) 23

(4) 181

Answer:

(2) 127

Hint:

316,18,112,118

a = 316, r = 18 ÷ 316 = 18 × 163 = 23

The next term is = 118 × 23 = 127

Question 14.

If the sequence t_{1},t_{2},t_{3} … are in A.P. then the sequence t_{6},t_{12},t_{18} … is

(1) a Geometric Progression

(2) an Arithmetic Progression

(3) neither an Arithmetic Progression nor a Geometric Progression

(4) a constant sequence

Answer:

(2) an Arithmetic Progression

Hint:

If t_{1}, t_{2}, t_{3}, … is 1, 2, 3, …

If t_{6} = 6, t_{12} = 12, t_{18} = 18 then 6, 12, 18 … is an arithmetic progression

Question 15.

The value of (1^{3} + 2^{3} + 3^{3} + ……. + 15^{3}) – (1 + 2 + 3 + …….. + 15) is …………….

(1) 14400

(2) 14200

(3) 14280

(4) 14520

Answer:

(3) 14280

Hint:

1202 – 120 = 120(120 – 1)

120 × 119 = 14280

**Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.9,**

Question 1.

Find the sum of the following series

(i) 1 + 2 + 3 + …….. + 60

(ii) 3 + 6 + 9 + …….. +96

(iii) 51 + 52 + 53 + …….. + 92

(iv) 1 + 4 + 9 + 16 + …….. + 225

(v) 6^{2} + 7^{2} + 8^{2} + …….. + 21^{2}

(vi) 10^{3} + 11^{3} + 12^{3} + …….. + 20^{3}

(vii) 1 + 3 + 5 + …… + 71

Solution:

(i) 1 + 2 + 3 + …….. + 60 = 60×612

[Using n(n+1)2 formula]

= 1830.

(ii) 3 + 6 + 9 + …….. + 96 = 3(1 + 2 + 3 + ……… + 32)

= 3×32×332

= 1584

(iii) 51 + 52 + 53 + …….. + 92 = (1 + 2 + 3 + ……. + 92) – (1 + 2 + 3 + …… + 50)

= 4278 – 1275

= 3003

(iv) 1 + 4 + 9 + 16 + …….. + 225 = 1^{2} + 2^{2} + 3^{2 }+ 4^{2} + ………… + 15^{2}

15×16×316

[using n(n+1)(2n+1)6] formula

= 1240

(v) 6^{2} + 7^{2} + 8^{2} + …….. + 21^{2} = 1 + 2^{2} + 3^{2} + 4^{2} + ………… + 21^{2} – (1 + 2^{2} + ………… + 5^{2})

= 21×22×436 – 5×6×116

= 3311 – 55

= 3256.

(vi) 10^{3} = 11^{3} + 12^{3} + …….. + 20^{3} = 1^{3} + 2^{3}+ 3^{3} + ………… + 20^{3} – (1^{3} + 2^{3} + 3^{3} + …………. + 9^{3})

[Using (n(n+1)2)^{2} formula]

= 210^{2} – 45^{2} = 44100 – 2025

= 42075

(vii) 1 + 3 + 5+ … + 71

Here a = 1; d = 3 – 1 = 2; l = 71

Question 2.

If 1 + 2 + 3 + …. + k = 325 , then find 1^{3} + 2^{3}+ 3^{3} + …………. + k^{3}

Answer:

1 + 2 + 3 + …. + k = 325

k(k+1)2 = 325 ……(1)

= 325^{2} (From 1)

= 105625.

Question 3.

If 1^{3} + 2^{3} + 3^{3} + ………… + K^{3} = 44100 then find 1 + 2 + 3 + ……. + k

Answer:

1^{3} + 2^{3} + 3^{3} + ………….. + k^{3} = 44100

k(k+1)2 = 44100−−−−−√ = 210

1 + 2 + 3 + …… + k = k(k+1)2

= 210

Question 4.

How many terms of the series 1^{3} + 2^{3} + 3^{3}+ …………… should be taken to get the sum 14400?

Answer:

1^{3} + 2^{3} + 3^{3} + ……. + n^{3} = 14400

n(n+1)2 = 14400−−−−−√

n(n+1)2 = 120 ⇒ n^{2} + n = 240

n^{2} + n – 240 = 0

(n + 16) (n – 15) = 0

(n + 16) = 0 or (n – 15) = 0

n = -16 or n = 15 (Negative will be omitted)

∴ The number of terms taken is 15

Question 5.

The sum of the squares of the first n natural numbers is 285, while the sum of their cubes is 2025. Find the value of n.

Answer:

1^{2} + 2^{2} + 3^{2} + …. + n^{2} = 285

Question 6.

Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm, …, 24 cm. How much area can be decorated with these colour papers?

Answer:

Area of 15 square colour papers

= 10^{2} + 11^{2} + 12^{2} + …. + 24^{2}

= (1^{2} + 2^{2} + 3^{2} + …. + 24^{2}) – (1^{2} + 2^{2} + 9^{2})

= 24×25×496−9×10×196

= 4 × 25 × 49 – 3 × 5 × 19

= 4900 – 285

= 4615

Area can be decorated is 4615 cm^{2}

Question 7.

Find the sum of the series (2^{3} – 1)+(4^{3} – 3^{3}) + (6^{3} – 15^{3}) + …….. to

(i) n terms

(ii) 8 terms

Answer:

Sum of the series = (2^{3} – 1) + (4^{3} – 3^{3}) + (6^{3} – 15^{3}) + …. n terms

= 2^{3} + 4^{3} + 6^{3} + …. n terms – (1^{3} + 3^{3} + 5^{3}+ …. n terms) …….(1)

2^{3} + 4^{3} + 6^{3} + …. n = ∑(2^{3} + 4^{3} + 6^{3} + ….(2n)^{3}]

∑ 2^{3} (1^{3} + 2^{3} + 3^{3} + …. n^{3})

= 8 (n(n+1)2)^{2}

= 2[n (n + 1)]^{2}

1^{3} + 3^{3} + 5^{3} + ……….(2n – 1)^{3} [sum of first 2n cubes – sum of first n even cubes]

Substituting (2) and (3) in (1)

Sum of the series = 2n^{2} (n + 1)^{2} – n^{2} (2n + 1)^{2} + 2n^{2}(n + 1)^{2}

= 4n^{2} (n + 1)^{2} – n^{2} (2n + 1)^{2}

= n^{2} [(4(n + 1)^{2} – (2n + 1)^{2}]

= n^{2} [4n^{2} + 4 + 8n – 4n^{2} – 1 – 4n]

= n^{2} [4n + 3]

= 4n^{3} + 3n^{2}

(ii) when n = 8 = 4(8)^{3} + 3(8)^{2}

= 4(512) + 3(64)

= 2240.

**Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.8,**

Question 1.

Which of the following sequences are in G.P?

(i) 3,9,27,81,…

(ii) 4,44,444,4444,…

(iii) 0.5,0.05,0.005,

(iv) 13,16,112, ………….

(v) 1, -5, 25,-125,…

(vi) 120, 60, 30, 18,…

(vii) 16, 4, 1, 14, ……….

Answer:

Question 3.

In a G.P. 729, 243, 81,… find t_{7}.

Answer:

The G.P. is 729, 243, 81,….

Question 4.

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

Solution:

G.P = x + 6, x + 12, x + 15

In G.P r = t2t1=t3t2

x+12x+6=x+15x+12

(x + 12)^{2} = (x + 6) (x + 5)

x^{2} + 24x + 144 = x^{2} + 6x + 15x + 90

24x – 21x = 90 – 144

3x = -54

x = −543 = -18

x = -18

Question 5.

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

(i) 4,8,16,…,8192?

Answer:

Here a = 4; r = 84 = 2

t_{n} = 8192

a . r^{n-1} = 8192 ⇒ 4 × 2^{n-1} = 8192

2^{n-1} = 81924 = 2048

2^{n-1} = 2^{11} ⇒ n – 1 = 11

n = 11 + 1 ⇒ n = 12

Number of terms = 12

(ii) 13, 19, 127, ……………, 12187

Answer:

a = 13 ; r = 19 ÷ 13 = 19 × 31 = 13

n – 1 = 6 ⇒ n = 6 + 1 = 7

Number of terms = 7

Question 6.

In a G.P. the 9^{th} term is 32805 and 6^{th} term is 1215. Find the 12^{th} term.

Answer:

Given, 9^{th} term = 32805

a. r^{n-1} = 12187

t_{9} = 32805 [t_{n} = ar^{n-1}]

a.r^{8} = 32805 …..(1)

6^{th} term = 1215

a.r^{5} = 1215 …..(2)

Divide (1) by (2)

ar8ar5 = 328051215 ⇒ r^{3} = 6561243

= 218781 = 72927 = 2439 = 813

r^{3} = 27 ⇒ r^{3} = 3^{3}

r = 3

Substitute the value of r = 3 in (2)

a. 3^{5} = 1215

a × 243 = 1215

a = 1215243 = 5

Here a = 5, r = 3, n = 12

t_{12} = 5 × 3^{(12-1)}

= 5 × 3^{11}

∴ 12^{th} term of a G.P. = 5 × 3^{11}

Question 7.

Find the 10th term of a G.P. whose 8^{th} term is 768 and the common ratio is 2.

Solution:

t_{8} = 768 = ar^{7}

r = 2

t_{10} = ar^{9} = ar^{7} × r × r

= 768 × 2 × 2 = 3072

Question 8.

If a, b, c are in A.P. then show that 3^{a}, 3^{b}, 3^{c} are in G.P.

Answer:

a, b, c are in A.P.

t_{2} – t_{1} = t_{3} – t_{2}

b – a = c – b

2b = a + c …..(1)

3^{a}, 3^{b}, 3^{c} are in G.P.

From (1) and (2) we get

3^{a}, 3^{b}, 3^{c} are in G.P.

Question 9.

In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is 572. Find the three terms.

Answer:

Let the three terms of the G.P. be ar, a, ar

Product of three terms = 27

ar × a × ar = 27

a^{3} = 27 ⇒ a^{3} = 3^{3}

a = 3

Sum of the product of two terms taken at a time is 572

6r^{2} – 13r + 6 = 0

6r^{2} – 9r – 4r + 6 = 0

3r (2r – 3) -2(2r – 3) = 0

(2r – 3) (3r – 2) = 0

2r – 3 = 0 or 3r – 2 = 0

2r = 3 (or) 3r – 2 = 0

r = 32 (or) r = 23

∴ The three terms are 2, 3 and 92 or 92, 3 and 2

Question 10.

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

Answer:

Starting salary (a) = ₹ 60000

Increased salary = 5% of starting salary

= 5100 × 60000

= ₹ 3000

Starting salary for the 2nd year = 60000 + 3000

= ₹ 63000

Year increase = 5% of 63000

= 5100 × 63000

= ₹ 3150

Starting salary for the 3^{rd} year = 63000 + 3150

= ₹ 66150

60000, 63000, 66150,…. form a G.P.

a = 60000; r = 6300060000 = 6360 = 2120

t_{n} = an^{n-1}

t_{5} = (60000) (2120)^{4}

= 60000 × 2120 × 2120 × 2120 × 2120

= 6×21×21×21×212×2×2×2

= 72930.38

5% increase = 5100 × 72930.38

= ₹ 3646.51

Salary after 5 years = ₹ 72930.38 + 3646.51

= ₹ 76576.90

= ₹ 76577

Question 11.

Sivamani is attending an interview for a job and the company gave two offers to him. Offer A: ₹20,000 to start with followed by a guaranteed annual increase of 6% for the first 5 years.

Offer B: ₹22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.

What is his salary in the 4^{th} year with respect to the offers A and B?

Answer:

Starting salary (a) = ₹ 20,000

Annual increase = 6% of 20000

= 5100 × 20000

= ₹ 1200

Salary for the 2nd year = ₹ 20000 + 1200

= ₹ 21200

Here a = 20,000; r = 2120020000 = 212200 = 106100 = 5350

n = 4 years

t_{n} = ar^{n-1}

Salary at the end of 4^{th} year = 23820

For B

Starting salary = ₹ 22000

(a) = 22000

Annual increase = 3% of 22000

= 3100 × 22000

= ₹ 660

Salary for the 2nd year = ₹ 22000 + ₹ 660

= ₹ 22,660

Here a = 22000; r = 2266022000

= 22662200 = 11331100 = 103100

Salary at the end of 4th year = 22000 × (103100)^{4-1}

= 22000 × (103100)^{3}

= 22000 × 103100 × 103100 × 103100

= 24039.99 = 24040

4^{th} year Salary for A = ₹ 23820 and 4^{th} year Salary for B = ₹ 24040

Question 12.

If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that x^{b-c} × y^{c-a} × z^{a-b} = 1

Answer:

a, b, c are three consecutive terms of an A.P

∴ a = a, b = a + dand c = a + 2d respectively ….(1)

x, y, z are three consecutive terms of a G.P

∴ x = x, y = xr, z = xr^{2} respective ……(2)

L.H.S = x^{b-c} × y^{c-a} × z^{a-b} ( Substitute the values from 1 and 2 we get)

L.H.S = R.H.S

Hence it is proved

**Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.7,**

Question 1.

Which of the following sequences are in G.P?

(i) 3,9,27,81,…

(ii) 4,44,444,4444,…

(iii) 0.5,0.05,0.005,

(iv) 13,16,112, ………….

(v) 1, -5, 25,-125,…

(vi) 120, 60, 30, 18,…

(vii) 16, 4, 1, 14, ……….

Answer:

Question 2.

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

(i) a = 6, r = 3

Answer:

a = 6, r = 3

ar = 6 × 3 = 18,

ar^{2} = 6 × 9 = 54

The three terms are 6, 18 and 54

(ii) a = 2–√, r = 2–√.

Answer:

ar = 2–√ × 2–√ = 2,

ar^{2} = 2–√ × 2 = 2 2–√

The three terms are 2–√, 2 and 22–√

(iii) a = 1000, r = 25

Answer:

ar = 1000 × 25 = 400,

ar^{2} = 1000 × 425 = 40 × 4 = 160

The three terms are 1000,400 and 160.

Question 3.

In a G.P. 729, 243, 81,… find t_{7}.

Answer:

The G.P. is 729, 243, 81,….

Question 4.

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

Solution:

G.P = x + 6, x + 12, x + 15

In G.P r = t2t1=t3t2

x+12x+6=x+15x+12

(x + 12)^{2} = (x + 6) (x + 5)

x^{2} + 24x + 144 = x^{2} + 6x + 15x + 90

24x – 21x = 90 – 144

3x = -54

x = −543 = -18

x = -18

Question 5.

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

(i) 4,8,16,…,8192?

Answer:

Here a = 4; r = 84 = 2

t_{n} = 8192

a . r^{n-1} = 8192 ⇒ 4 × 2^{n-1} = 8192

2^{n-1} = 81924 = 2048

2^{n-1} = 2^{11} ⇒ n – 1 = 11

n = 11 + 1 ⇒ n = 12

Number of terms = 12

(ii) 13, 19, 127, ……………, 12187

Answer:

a = 13 ; r = 19 ÷ 13 = 19 × 31 = 1

t_{n} = 12187

a. r^{n-1} = 12187

n – 1 = 6 ⇒ n = 6 + 1 = 7

Number of terms = 7

Question 6.

In a G.P. the 9^{th} term is 32805 and 6^{th} term is 1215. Find the 12^{th} term.

Answer:

Given, 9^{th} term = 32805

a. r^{n-1} = 12187

t_{9} = 32805 [t_{n} = ar^{n-1}]

a.r^{8} = 32805 …..(1)

6^{th} term = 1215

a.r^{5} = 1215 …..(2)

Divide (1) by (2)

ar8ar5 = 328051215 ⇒ r^{3} = 6561243

= 218781 = 72927 = 2439 = 813

r^{3} = 27 ⇒ r^{3} = 3^{3}

r = 3

Substitute the value of r = 3 in (2)

a. 3^{5} = 1215

a × 243 = 1215

a = 1215243 = 5

Here a = 5, r = 3, n = 12

t_{12} = 5 × 3^{(12-1)}

= 5 × 3^{11}

∴ 12^{th} term of a G.P. = 5 × 3^{11}

Question 7.

Find the 10th term of a G.P. whose 8^{th} term is 768 and the common ratio is 2.

Solution:

t_{8} = 768 = ar^{7}

r = 2

t_{10} = ar^{9} = ar^{7} × r × r

= 768 × 2 × 2 = 3072

Question 8.

If a, b, c are in A.P. then show that 3^{a}, 3^{b}, 3^{c} are in G.P.

Answer:

a, b, c are in A.P.

t_{2} – t_{1} = t_{3} – t_{2}

b – a = c – b

2b = a + c …..(1)

3^{a}, 3^{b}, 3^{c} are in G.P.

From (1) and (2) we get

3^{a}, 3^{b}, 3^{c} are in G.P.

Question 9.

In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is 572. Find the three terms.

**Answer:**

Let the three terms of the G.P. be ar, a, ar

Product of three terms = 27

ar × a × ar = 27

a^{3} = 27 ⇒ a^{3} = 3^{3}

a = 3

Sum of the product of two terms taken at a time is 572

6r^{2} – 13r + 6 = 0

6r^{2} – 9r – 4r + 6 = 0

3r (2r – 3) -2(2r – 3) = 0

(2r – 3) (3r – 2) = 0

2r – 3 = 0 or 3r – 2 = 0

2r = 3 (or) 3r – 2 = 0

r = 32 (or) r = 23

∴ The three terms are 2, 3 and 92 or 92, 3 and 2

**Question 10.**

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

**Answer:**

Starting salary (a) = ₹ 60000

Increased salary = 5% of starting salary

= 5100 × 60000

= ₹ 3000

Starting salary for the 2nd year = 60000 + 3000

= ₹ 63000

Year increase = 5% of 63000

= 5100 × 63000

= ₹ 3150

Starting salary for the 3^{rd} year = 63000 + 3150

= ₹ 66150

60000, 63000, 66150,…. form a G.P.

a = 60000; r = 6300060000 = 6360 = 2120

t_{n} = an^{n-1}

t_{5} = (60000) (2120)^{4}

= 60000 × 2120 × 2120 × 2120 × 2120

= 6×21×21×21×212×2×2×2

= 72930.38

5% increase = 5100 × 72930.38

= ₹ 3646.51

Salary after 5 years = ₹ 72930.38 + 3646.51

= ₹ 76576.90

= ₹ 76577

**Question 11.**

Sivamani is attending an interview for a job and the company gave two offers to him.

**Offer A: **₹20,000 to start with followed by a guaranteed annual increase of 6% for the first 5 years.**Offer B:** ₹22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.

What is his salary in the 4^{th} year with respect to the offers A and B?

**Answer:**

Starting salary (a) = ₹ 20,000

Annual increase = 6% of 20000

= 5100 × 20000

= ₹ 1200

Salary for the 2nd year = ₹ 20000 + 1200

= ₹ 21200

Here a = 20,000; r = 2120020000 = 212200 = 106100 = 5350

n = 4 years

t_{n} = ar^{n-1}

Salary at the end of 4^{th} year = 23820

For B

Starting salary = ₹ 22000

(a) = 22000

Annual increase = 3% of 22000

= 3100 × 22000

= ₹ 660

Salary for the 2nd year = ₹ 22000 + ₹ 660

= ₹ 22,660

Here a = 22000; r = 2266022000

= 22662200 = 11331100 = 103100

Salary at the end of 4th year = 22000 × (103100)^{4-1}

= 22000 × (103100)^{3}

= 22000 × 103100 × 103100 × 103100

= 24039.99 = 24040

4^{th} year Salary for A = ₹ 23820 and 4^{th} year Salary for B = ₹ 24040

Question 12.

If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that x^{b-c} × y^{c-a} × z^{a-b} = 1

Answer:

a, b, c are three consecutive terms of an A.P

∴ a = a, b = a + dand c = a + 2d respectively ….(1)

x, y, z are three consecutive terms of a G.P

∴ x = x, y = xr, z = xr^{2} respective ……(2)

L.H.S = x^{b-c} × y^{c-a} × z^{a-b} ( Substitute the values from 1 and 2 we get)

L.H.S = R.H.S

Hence it is proved

**Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.6,**

Question 1.

Find the sum of the following

(i) 3, 7, 11,… up to 40 terms.

Answer:

3,7,11,… up to 40 terms

First term (a) = 3

Common difference (d) = 7 – 3 = 4

Number of terms (n) = 40

S_{n} = n2 [2a + (n – 1) d]

S_{40} = 402 [6 + 39 × 4] = 20 [6 + 156]

= 20 × 162

S_{40} = 3240

(ii) 102,97, 92,… up to 27 terms.

Answer:

Here a = 102, d = 97 – 102 = -5

n = 27

S_{n} = n2 [2a + (n – 1)d]

S_{27} = 272 [2(102) + 26(-5)]

= 272 [204 – 130]

= 272 × 74

= 27 × 37 = 999

S_{27} = 999

(iii) 6 + 13 + 20 + …. + 97

Answer:

Here a = 6, d = 13 – 6 = 7, l = 97

n = l−ad + 1

= 97−67 + 1

= 917 + 1 =

13 + 1 = 14

S_{n} = n2 (a + l)

S_{n} = 142 (a + l)

S_{n} = 142 (6 + 97)

= 7 × 103

S_{n} = 721

Question 2.

How many consecutive odd integers beginning with 5 will sum to 480?

Answer:

First term (a) = 5

Common difference (d) = 2

(consecutive odd integer)

S_{n} = 480

n2 [2a + (n-1) d] = 480

n2 [10 + (n-1)2] = 480

n + 24 = 0 or n – 20 = 0

n = -24 or n = 20

[number of terms cannot be negative]

∴ Number of consecutive odd integers is 20

Question 3.

Find the sum of first 28 terms of an A.P. whose n^{th} term is 4n – 3.

Solution:

n = 28

t_{n} = 4n – 3

t_{1} = 4 × 1 – 3 = 1

t_{2} = 4 × 2 – 3 = 5

t_{28} = 4 × 28 – 3

= 112 – 3 = 109

∴ a = 1, d = t_{2} – t_{1} = 5 – 1 = 4

l = 109.

S_{n} = n2 (2a+(n – 1)d)

S_{28} = 282 (2 × 1 + 27 × 4)

= 14(2 + 108)

= 14 × 110

= 1540

Question 4.

The sum of first n terms of a certain series is given as 2n^{2} – 3n . Show that the series is an A.P.

Answer:

Let tn be nth term of an A.P.

t_{n} = Sn – S_{n-1}

= 2n^{2} – 3n – [2(n – 1)^{2} – 3(n – 1)]

= 2n^{2} – 3n – [2(n^{2} – 2n + 1) – 3n + 3]

= 2n^{2} – 3n – [2n^{2} – 4n + 2 – 3n + 3]

= 2n^{2} – 3n – [2n^{2} – 7n + 5]

= 2n^{2} – 3n – 2n^{2} + 7n – 5

t_{n} = 4n – 5

t_{1} = 4(1) – 5 = 4 – 5 = -1

t_{2} = 4(2) -5 = 8 – 5 = 3

t_{3} = 4(3) – 5 = 12 – 5 = 7

t_{4} = 4(4) – 5 = 16 – 5 = 11

The A.P. is -1, 3, 7, 11,….

The common difference is 4

∴ The series is an A.P.

Question 5.

The 104th term and 4^{th} term of an A.P are 125 and 0. Find the sum of first 35 terms?

Answer:

104th term of an A.P = 125

t_{104} = 125

[t_{n} = a + (n – 1) d]

a + 103d = 125 …..(1)

4^{th} term = 0

t_{4} = 0

a + 3d = 0 …..(2)

Sum of 35 terms = 612.5

Question 6.

Find the sum of ail odd positive integers less than 450.

Answer:

Sum of odd positive integer less than 450

1 + 3 + 5 + …. 449

Here a = 1, d = 3 – 1 = 2,l = 449

Aliter: Sum of all the positive odd intergers

= n^{2}

= 225 × 225

= 50625

Sum of the odd integers less than 450

= 50625

**Question 7.**

Find the sum of all natural numbers between 602 and 902 which are not divisible by 4?

Answer:

First find the sum of all the natural’s number between 602 and 902

Here a = 603, d = 1, l = 901

Find the sum of all the numbers between 602 and 902 which are divisible by 4.

Here a = 604; l = 900; d = 4

Sum of the numbers which are not divisible

by 4 = S_{n1} – S_{n2}

= 224848 – 56400

= 168448

Sum of the numbers = 168448

**Question 8.**

Raghu wish to buy a laptop. He can buy it by paying ₹40,000 cash or by giving it in 10 installments as ₹4800 in the first month, ₹4750 in the second month, ₹4700 in the third month and so on. If he pays the money in this fashion, find

(i) total amount paid in 10 installments.

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

Solution:

4800 + 4750 + 4700 + … 10 terms

Here a = 4800

(i) d = t_{2} – t_{1} = 4750 – 4800 = -50

n = 10

S_{n} = n2 (2a + (n – 1)d)

S_{10} = 102 (2 × 4800 + 9 × -50)

= 5 (9600 – 450)

= 5 × 9150 = 45750

Total amount paid in 10 installments = ₹ 45750.

(ii) The extra amount he pays in installments

= ₹ 45750 – ₹ 40,000

= ₹ 5750

**Question 9.**

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

Answer:

(i) Total loan amount = ₹ 65,000

S_{n} = 65,000

First month payment (a) = 400

Every month increasing ₹ 300

d = 300

S_{n} = n2 [2a + (n-1)d]

65000 = n2 [2(400) + (n – 1)300]

130000 = n [800 + 300n – 300]

= n [500 + 300n]

13000 = 500n + 300n^{2}

Dividing by (100)

Number of installments will not be negative

∴ Time taken to pay the loan is 20 months.

**Question 10.**

A brick staircase has a total of 30 steps. The bottom step requires 100 bricks. Each successive step requires two bricks less than the previous step.

(i) How many bricks are required for the top most step?

(ii) How many bricks are required to build the stair case?

Solution:

100 + 98 + 96 + 94 + … 30 steps.

Here

a = 100

d = -2

n = 30

∴ S_{n} = n2 (2a + (n – 1)d)

S_{30} = 302 (2 × 100 + 29 × -2)

= 15(200 – 58)

= 15 × 142

= 2130

t_{30} = a + (n – 1)d

= 100 + 29 × -2

= 100 – 58

= 42

(i) No. of bricks required for the top step are 42.

(ii) No. of bricks required to build the stair case are 2130.

**Question 11.**

If S_{1}, S_{2} , S_{3}, ….S_{m} are the sums of n terms of m A.P.,s whose first terms are 1,2, 3…… m and whose common differences are 1,3,5,…. (2m – 1) respectively, then show that (S_{1} + S_{2} + S_{3} + ……. + S_{m}) = 12 mn(mn + 1)

Answer:

First terms of an A.P are 1, 2, 3,…. m

The common difference are 1, 3, 5,….

(2m – 1)

By adding (1) (2) (3) we get

S_{1} + S_{2} + S_{3} + …… + S_{m} = n2 (n + 1) + n2 (3n + 1) + n2 (5n + 1) + ….. + n2 [n(2m – 1 + 1)]

= n2 [n + 1 + 3n + 1 + 5n + 1 ……. + n (2m – 1) + m)]

= n2 [n + 3n + 5n + ……. n(2m – 1) + m]

= n2 [n (1 + 3 + 5 + ……(2m – 1)) + m

= n2 [n(m2) (2m) + m]

= n2 [nm^{2} + m]

S_{1} + S_{2} + S_{3} + ……….. + S_{m} = mn2 [mn + 1]

Hint:

1 + 3 + 5 + ……. + 2m – 1

S_{n} = n2 (a + l)

= m2 (1 + 2m -1)

= m2 (2m)

Question 12.

Find the sum

Answer:

**Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.5,**

Question 1.

Check whether the following sequences are in A.P.?

(i) a – 3, a – 5, a – 7,…

Answer:

a – 3, a – 5, a – 7…….

t_{2} – t_{1} = a – 5 – (a – 3)

= a – 5 – a + 3

= -2

t_{3} – t_{2} = a – 7 – (a – 5)

= a – 7 – a + 5

= -2

t_{2} – t_{1} = t_{3} – t_{2}

(common difference is same)

The sequence is in A.P.

(ii) 12, 13, 14, 15, ……….

Answer:

t_{2} – t_{1} = 13 – 12 = 2−36 = −16

t_{3} – t_{2} = 14 – 13 = 3−412 = −112

t_{2} – t_{1} ≠ t_{3} – t_{2}

The sequence is not in A.P.

(iii) 9, 13, 17, 21, 25,…

Answer:

t_{2} – t_{1} = 13 – 9 = 4

t_{3} – t_{2} = 17 – 13 = 4

t_{4} – t_{3} = 21 – 17 = 4

t_{5} – t_{4} = 25 – 21 = 4

Common difference are equal

∴ The sequence is in A.P.

(iv) −13, 0, 13, 23

t_{2} – t_{1} = 0 – (-13)

= 0 + 13 = 13

t_{3} – t_{2} = 13 – 0 = 13

t_{2} – t_{1} = t_{3} – t_{2}

The sequence is in A.P.

(v) 1,-1, 1,-1, 1, -1, …

t_{2} – t_{1} = -1 – 1 = -2

t_{3} – t_{2} = 1 – (-1) = 1 + 1 = 2

t_{4} – t_{3} = -1-(1) = – 1 – 1 = – 2

t_{5} – t_{4} = 1 – (-1) = 1 + 1 = 2

Common difference are not equal

∴ The sequence is not an A.P.

Question 2.

First term a and common difference d are given below. Find the corresponding A.P. ?

(i) a = 5 ,d = 6

Answer:

Here a = 5,d = 6

The general form of the A.P is a, a + d, a + 2d, a + 3d….

The A.P. 5, 11, 17, 23 ….

(ii) a = 7, d = -5

Answer:

The general form of the A.P is a, a + d,

a + 2d, a + 3d… .

The A.P. 7, 2, -3, -8 ….

(iii) a = 34, d = 12

Answer:

The general form of the A.P is a, a + d, a + 2d, a + 3d….

34,34 + 12,34 + 2(12), 34 + 3 (12)

The A.P. 34, 54, 74, …….

Question 3.

Find the first term and common difference of the Arithmetic Progressions whose n^{th} terms are given below

(i) t_{n} = -3 + 2n

(ii) t_{n} = 4 – 7n

Solution:

(i) a = t_{1} = -3 + 2(1) = -3 + 2 = -1

d = t_{2} – t_{1}

Here t_{2} = -3 + 2(2) = -3 + 4 = 1

∴ d = t_{2} – t_{1} = 1 – (-1) = 2

(ii) a = t_{1} = 4 – 7(1) = 4 – 7 = -3

d = t_{2} – t_{1}

Here t_{2} = 4 – 7(2) = 4 – 14 – 10

∴ d = t_{2} – t_{1} = 10 – (-3) = -7

Question 4.

Find the 19^{th} term of an A.P. -11, -15, -19,…

Answer:

First term (a) = -11

Common difference (d) = -15 -(-11)

= -15 + 11 = -4

n = 19

t_{n} = a + (n – 1) d

t_{n} = -11 + 18(-4)

= -11 – 72

t_{19} = -83

19^{th} term of an A.P. is – 83

Question 5.

Which term of an A.P. 16, 11, 6,1, ……….. is -54?

Solution:

A.P = 16, 11,6, 1, ………..

It is given that

t_{n} = -54

a = 16, d = t_{2} – t_{1} = 11 – 16 = -5

∴ t_{n} = a + (n – 1)d

-54 = 16 + (n – 1) (-5)

-54 = 16 – 5n + 5

21 – 5n = -54

-5n = -54 -21

-5n = -75

n = 755 =15

∴ 15th term is -54.

Question 6.

Find the middle term(s) of an A.P. 9, 15, 21, 27, …, 183.

Answer:

First term (a) = 9

Last term (l) = 183

Common difference (d) = 15 – 9 = 6

n = l−ad + 1

= 183−96 + 1

= 1746 + 1

= 29 + 1

= 30

middle term = 15^{th} term of

16^{th} term

t_{n} = a + (n – 1)d

t_{15} = 9 + 14(6)

= 9 + 84 = 93

t_{16} = 9 + 15(6)

= 9 + 90 = 99

The middle term is 93 or 99

Question 7.

If nine times ninth term is equal to the fifteen times fifteenth term, show that six times twenty fourth term is zero.

Solution:

Nine times ninth term = Fifteen times fifteenth term

9t_{9} = 15t_{15}

9(a + 8d) = 5(a + 14d)

9a + 72d = 15a + 210

15a + 210d – 9a – 72d = 0

⇒ 6a + 138 d = 0

⇒ 6(a + 23 d) = 0

⇒ 6(a + (24 – 1)d) = 0

⇒ 6t_{24} = 0. Hence it is proved.

Question 8.

If 3 + k, 18 – k, 5k + 1 are in A.P. then find k?

Answer:

3 + k, 18 – k, 5k + 1 are in AP

∴ t_{2} – t_{1} = t_{3} – t_{2} (common difference is same)

18 – k – (3 + k) = 5k + 1 – (18 – k)

18 – k – 3 – k = 5k + 1 – 18 + k

15 – 2k = 6k – 17

32 = 8k

k = 328 = 4

The value of k = 4

Question 9.

Find x, y and z, given that the numbers x, 10, y, 24, z are in A.P.

Solution:

A.P = x, 10, y, 24, z,…

d = t_{2} – t_{1} = 10 – x ………….. (1)

= t_{3} – t_{2} = y – 10 ………….. (2)

= t_{4} – t_{3} = 24 – y …………. (3)

= t_{5} – t_{4} = z – 24 ………….. (4)

(2) and (3)

⇒ y – 10 = 24 – y

2y = 24 + 10 = 34

y = 342 = 17

(1) and (2)

⇒ 10 – x = y – 10

10 – x = 17 – 10 = 7

-x = 7 – 10

-x = -3 ⇒ x = 3

From (3) and (4)

24 – y = z – 24

24 – 17 = z – 24

7 = z – 24

∴ z = 7 + 24 = 31

∴ Solutions x = 3

y = 17

z = 31

Question 10.

In a theatre, there are 20 seats in the front row and 30 rows were allotted. Each successive row contains two additional seats than its front row. How many seats are there in the last row?

Answer:

Number of seats in the first row

(a) = 20

∴ t_{1} = 20

Number of seats in the second row

(t_{2}) = 20 + 2

= 22

Number of seats in the third row

(t_{3}) = 22 + 2

= 24

Here a = 20 ; d = 2

Number of rows

(n) = 30

t_{n} = a + (n – 1)d

t_{30} = 20 + 29(2)

= 20 + 58

t_{30} = 78

Number of seats in the last row is 78

Question 11.

The sum of three consecutive terms that are in A.P. is 27 and their product is 288. Find the three terms.

Solution:

Let the three consecutive terms be a – d, a, a + d

Their sum = a – d + a + a + d = 27

3a = 27

a = 273 = 9

Their product = (a – d)(a)(a + d) = 288

= 9(a^{2} – d^{2}) = 288

⇒ 9(9 – d^{2}) = 288

⇒ 9(81 – d^{2}) = 288

81 – d^{2} = 32

-d^{2} = 32 – 81

d^{2} = 49

⇒ d = ± 7

∴ The three terms are if a = 9, d = 7

a – d, a , a + d = 9 – 7, 9 + 7

A.P. = 2, 9, 16

if a = 9, d = -7

A.P. = 9 – (-7), 9, 9 + (-7)

= 16, 9, 2

Question 12.

The ratio of 6^{th} and 8^{th} term of an A.P is 7:9. Find the ratio of 9^{th} term to 13^{th} term.

Answer:

Given : t_{6} : t_{8} = 7 : 9 (using t_{n} = a + (n – 1)d

a + 5d : a + 7d = 7 : 9

9 (a + 5 d) = 7 (a + 7d)

9a + 45 d = 7a + 49d

9a – 7a = 49d – 45d

2a = 4d

a = 2d

To find t_{9} : t_{13}

t_{9} : t_{13} = a + 8d : a + 12d

= 2d + 8d : 2d + 12d

= 10d : 14d

= 5 : 7

∴ t_{9} : t_{13} = 5 : 7

Question 13.

In a winter season, the temperature of Ooty from Monday to Friday to be in A.P. The sum of temperatures from Monday to Wednesday is 0° C and the sum of the temperatures from Wednesday to Friday is 18° C. Find the temperature on each of the five days.

Answer:

Let the five days temperature be

(a – 2d), (a – d), a,(a + d) and (a + 2d)

Sum of first three days temperature = 0

a – 2d + a – d + a = 0

3a – 3d = 0

a – d = 0 …..(1)

Sum of the last three days temperature = 18°C

a + a + d + a + 2d = 18

3a + 3d = 18

(÷ by 3) ⇒ a + d = 6 ……(2)

By adding (1) and (2)

Substitute to value of a = 3 in (2)

d = 3

The temperature in 5 days are

(3 – 6), (3 – 3), 3, (3 + 3) and (3 + 6)

-3°C, 0°C, 3°C, 6°C, 9°C

Question 14.

Priya earned ₹15,000 in the first month. Thereafter her salary increases by ₹1500 per year. Her expenses are ₹13,000 during the first year and the expenses increases by ₹900 per year. How long will it take her to save ₹20,000 per month.

Answer:

Tabulate the given table

Monthly savings form an A.P.

2000, 2600, 3200 …..

a = 2000; d = 2600 – 2000 = 600

Given t_{n} = 20,000

t_{n} = a + (n – 1) d

20000 = 2000 + (n – 1) 600

20000 = 2000 + 600n – 600

= 1400 + 600n

20000 – 1400 = 600n

18600 = 600n

n = 18600600 = 31

He will take 31 years to save ₹ 20,000 per month.

**Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.4,**

Question 1.

Find the next three terms of the following sequence.

(i) 8, 24, 72,…

(ii) 5, 1, -3, …

(iii) 14, 29, 316

(i) 216, 648, 1944 (This sequence is multiple of 3)

Next three terms are 216, 648, 1944

(ii) Next three terms are -7, -11, -15

(adding -4 with each term)

(iii) Next three terms are 425,536 and 649

[using n(n+1)2]

Question 2.

Find the first four terms of the sequences whose nth terms are given by

(i) a_{n} = n^{3} – 2

(ii) a_{n} = (-1)^{n+1} n(n+1)

(iii) a_{n} = 2n^{2} – 6

Solution:

t_{n }= a_{n }= n^{3 }-2

(i) a_{1} = 1^{3} – 2 = 1 – 2 – 1

a_{2} = 2^{3} – 2 = 8 – 2 = 6

a_{3} = 3^{3} – 2 = 27 – 2 = 25

a_{4} = 4^{3} – 2 = 64 – 2 = 62

∴ The first four terms are -1, 6, 25, 62, ……….

(ii) a_{n} = (-1)^{n+1} n(n + 1)

a_{1} = (-1)^{1+1} (1) (1 +1)

= (-1)^{2} (1) (2) = 2

a_{2} = (-1)^{2+1} (2) (2 + 1)

= (-1)^{3} (2) (3)= -6

a_{3} = (-1)^{3+1} (3) (3 + 1)

= (-1)^{4} (3) (4) = 12

a_{4} = (-1)^{4+1} (4) (4 + 1)

= (-1)^{5} (4) (5) = -20

∴ The first four terms are 2, -6, 12, -20,…

(iii) a_{n} = 2n^{2} – 6

a_{1} = 2(1)^{2} – 6 = 2 – 6 = -4

a_{2} = 2(2)^{2} – 6 = 8 – 6 = 2

a_{3} = 2(3)^{2} – 6 = 18 – 6 = 12

a_{4} = 2(4)^{2} – 6 = 32 – 6 = 26

∴ The first four terms are -4, 2, 12, 26, …

Question 3.

Find the n^{th} term of the following sequences

(i) 2, 5, 10, 17, ……

Answer:

(1^{2} + 1);(2^{2} + 1),(3^{2} + 1),(4^{2} + 1)….

n^{th} term is n^{2} + 1

a_{n} = n^{2} + 1

(ii) 0,12,23 ……

Answer:

(1−11), (2−12), (3−13) …..

n^{th} term is n−1n

a_{n} = n−1n

(iii) 3,8,13,18,…….

Answer:

[5(1) -2], [5(2) – 2], [5(3) – 2], [5(4) – 2] ….

The n^{th} term is 5n – 2

a_{n} = 5n – 2

Question 4.

Find the indicated terms of the sequences whose n^{th} terms are given by

(i) a_{n} = 5nn+2 ; a_{6} and a_{13}

Answer:

a_{n} = 5nn+2

a_{6} = 5(6)6+2 = 308 = 154

a_{13} = 5(13)13+2 = 5×1315 = 133

a_{6} = 154, a_{13} = 133

(ii) a_{n} = – (n^{2} – 4); a_{4} and a_{11}

Answer:

a_{n} = -(n^{2} – 4)

a_{4} = -(4^{2} – 4)

= – (16 – 4)

= -12

a_{11} = -(11^{2} – 4)

= – (121 – 4)

= – 117

a_{4} = -12 and a_{11} = -117

Question 5.

Find a_{8} and a_{15} whose n^{th} term is a_{n}Answer:

Question 6.

If a_{1} = 1, a_{2} = 1 and a_{n} = 2a_{n-1} + a_{n-2}, n > 3, n ∈ N, then find the first six terms of the sequence.

Solution:

a_{1} = 1, a_{2} = 1, a_{n} = 2a_{n-1} + a_{n-2}

a_{3} = 2a_{(3-1)} + a_{(3-2)}

= 2a_{2} + a_{1}

= 2 × 1 + 1 = 3

a_{4} = 2a_{(4-1)} + a_{(4-2)}

= 2a_{3} + a_{2}

= 2 × 3 + 1 = 7

a_{5} = 2a_{(5-1)} + a_{(5-2)}

= 2a_{4} + a_{3}

= 2 × 7 + 3 = 17

a_{6} = 2a_{(6-1)} + a_{(6-2)}

= 2a_{5} + a_{4}

= 2 × 17 + 7

= 34 + 7

= 41

∴ The first six terms of the sequence are 1, 1, 3, 7, 17, 41 ………..

Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.3,

Question 1.

Find the least positive value of x such that

(i) 71 = x (mod 8)

Answer:

71 = 7 (mod 8)

∴ The value of x = 7

(ii) 78 + x = 3 (mod 5)

78 + x – 3 = 5n (n is any integer)

75 + x = 5n

(Let us take x = 5)

75 + 5 = 80 (80 is a multiple of 5)

∴ The least value of x is 5

(iii) 89 = (x + 3) (mod 4)

89 – (x + 3) = 4n

(n may be any integer)

89 – x – 3 = 4n

89 – x = 4n

86 – x is a multiple of 4

(84 is a multiple of 4)

86 – 2 = 4n

84 = 4n

The value of x is 2

(iv) 96 = x7 (mod 5)

96 – x7 = 5n (n may be any integer)

672 – x = 35n (multiple of 35 is 665)

672 – 7 = 665

∴ The value of x = 7

(v) 5x = 4 (mod 6)

5x – 4 = 6n (n may be any integer)

5x = 6n + 4

x = 6n+45

Substitute the value of n as 1, 6, 11, 16 …. as n values in x = 6n+45 which is divisible by 5.

2, 8, 14, 20,…………

The least positive value is 2.

Question 2.

If x is congruent to 13 modulo 17 then 7x -3 is congruent to which number modulo 17?

Solution:

x ≡ 13 (mod 17)

Let p be the required number …………. (1)

7x – 3 ≡ p (mod 17) ………….. (2)

From (1),

x – 13 = 17n for some integer M.

x – 13 is a multiple of 17.

x must be 30.

∴ 30 – 13 = 17

which is a multiple of 17.

From (2),

7 × 30 – 3 ≡ p (mod 17)

210 – 3 ≡ p (mod 17)

207 ≡ p (mod 17)

207 ≡ 3 (mod 17)

∴ P ≡ 3

Question 3.

Solve 5x ≡ 4 (mod 6)

5x – 4 = 6n (n may be any integer)

5x = 6n + 4

x = 6n+45

The value of n 1, 6, 11, 16 ……..

∴ The value of x is 2, 8, 14, 20 …………..

Question 4.

Solve 3x – 2 = 0 (mod 11)

Answer:

Given 3x – 2 = 0(mod 11)

3x – 2 = 11n (n may be any integer)

3x = 2 + 11n

x = 11n+23

Substitute the value of n = 2, 5, 8, 11 ….

When n ≡ 2 ⇒ x = 22+23 = 243 = 8

When n = 5 ⇒ x = 55+23 = 573 = 19

When n = 8 ⇒ x = 88+23 = 903 = 30

When n = 11 ⇒ x = 121+23 = 1233 = 41

∴ The value of x is 8, 19, 30,41

Question 5.

What is the time 100 hours after 7 a.m.?

Answer:

100 ≡ x (mod 12) Note: In a clock every 12 hours

100 ≡ 4 (mod 12) the numbers repeats.

The time repeat after 7 am is 7 + 4 = 11 o’ clock (or) 11 am.

Question 6.

What is time 15 hours before 11 p.m.?

Solution:

15 ≡ x (mod 12)

15 – x = 12n

15 – x is a multiple of 12 x must be 3.

∴ The time 15 hrs before 11 O’clock is 11 – 3 = 8 O’ clock i.e. 8 p.m

Question 7.

Today is Tuesday. My uncle will come after 45 days. In which day my uncle will be coming?

Answer:

Number of days in a week = 7

45 ≡ x (mod 7)

45 ≡ 3 (mod 7)

The value of x must be 3.

Three days after tuesday is friday uncle will come on friday.

Question 8.

Prove that 2^{n} + 6 × 9^{n} is always divisible by 7 for any positive integer n.

Solution:

2^{1} + 6 × 9^{1} = 2 + 54 = 56 is divisible by 7

When n = k,

2^{k} + 6 × 9^{k} = 7 m [where m is a scalar]

⇒ 6 × 9^{k} = 7 m – 2^{k} …………. (1)

Let us prove for n = k + 1

Consider 2^{k+1} + 6 × 9^{k+1} = 2^{k+1} + 6 × 9^{k} × 9

= 2^{k+1} + (7m – 2^{k})9 (using (1))

= 2^{k+1} + 63m – 9.2^{k} = 63m + 2^{k}.2^{1} – 9.2^{k}

= 63m – 2^{k} (9 – 2) = 63m – 7.2^{k}

= 7 (9m – 2^{k}) which is divisible by 7

∴ 2^{n} + 6 × 9^{n} is divisible by 7 for any positive integer n

Question 9.

Find the remainder when 2^{81} is divided by 17?

Answer:

2^{81} ≡ x(mod 17)

2^{40} × 2^{40} × 2^{1} ≡ x(mod 17)

(2^{4})^{10} × (2^{4})^{10} × 2^{1} ≡ x(mod 17)

(16)^{10} × (16)^{10} × 2^{1} ≡ x(mod 17)

(16^{2})^{5} × (16^{2})^{5} × 2^{1} ≡ x(mod 17)

= 1 × 1 × 2 (mod 17)

[(16)^{2} = 256 = 1 (mod 17)]

= 2 (mod 17)

2^{81} = 2(mod 17)

∴ x = 2

The remainder is 2

Question 10.

The duration of flight travel from Chennai to London through British Airlines is approximately 11 hours. The airplane begins its journey on Sunday at 23:30 hours. If the time at Chennai is four and half hours ahead to that of London’s time, then find the time at London, when will the flight lands at London Airport?

Answer:

Duration of the flight time = 11 hours

(Chennai to London)

Starting time on Sunday = 23 : 30 hour

Time difference is 4 12 horns ahead to london

The time to reach London airport = (10.30 – 4.30)

= 6 am

The first reach the london airport next day (monday) at 6 am

**Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.2,**

Question 1.

For what values of natural number n, 4th can end with the digit 6?

Answer:

4n = (2^{2})^{n} = 2^{2n}

= 2^{n} × 2^{n}

2 is a factor of 4^{n}

∴ 4^{n} is always even.

Question 2.

If m, n are natural numbers, for what values of m, does 2^{n} × 5^{n} ends in 5?

Solution:

2^{n} × 5^{m}

2^{n} is always even for all values of n.

5^{m} is always odd and ends with 5 for all values of m.

But 2^{n} × 5^{m} is always even and ends in 0.

∴ 2^{n} × 5^{m} cannot end with the digit 5 for any values of m. No value of m will satisfy 2^{n} × 5^{m} ends in 5.

Question 3.

Find the H.C.F. of 252525 and 363636.

Answer:

To find the HCF of 252525 and 363636 by using Euclid’s Division algorithm.

363636 = 252525 × 1 + 111111

The remainder 111111 ≠ 0

By division of Euclid’s algorithm

252525 = 111111 × 2 + 30303

The remainder 30303 ≠ 0

Again by division of Euclid’s algorithm

111111 = 30303 × 3 + 20202

The remainder 20202 ≠ 0

Again by division of Euclid’s algorithm.

30303 = 20202 + 10101

The remainder 10101 ≠ 0

Again by division of Euclid’s algorithm.

20202 = 10101 × 2 + 0

The remainder is 0

∴ The H.C.F. is 10101

Question 4.

If 13824 = 2^{a} × 3^{b} then find a and b?

Answer:

Using factor tree method factorise 13824

13824 = 2^{9} × 3^{3}

Given 13824 = 2a × 3b

Compare we get a = 9 and b = 3

Aliter:

13824 = 2^{9} × 3^{3}

Compare with

13824 = 2^{a} × 3^{b}

The value of a = 9 b = 3

Question 5.

If p_{1}^{x1} × p_{2}^{x2} × p_{3}^{x3} × p_{4}^{x4} = 113400 where p_{1} p_{2}, p_{3}, p_{4} are primes in ascending order and x_{1}, x_{2}, x_{3}, x_{4}, are integers, find the value of p_{1},p_{2},p_{3},p_{4} and x_{1},x_{2},x_{3},x_{4}.

Answer:

Given 113400 = p_{1}^{x1} × p_{2}^{x2} × p_{3}^{x3} × p_{4}^{x4}

Using tree method factorize 113400

113400 = 23 × 34 × 52 × 7

compare with

113400 = p_{1}^{x1} × p_{2}^{x2} × p_{3}^{x3} × p_{4}^{x4}

P_{1} = 2, x_{1} = 3

P_{2} = 3, x_{2} = 4

P_{3} = 5, x_{3} = 2

P_{4} = 7, x_{4} = 1

Question 6.

Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of Arithmetic.

Answer:

Factorise 408 and 170 by factor tree method

Question 7.

Find the greatest number consisting of 6 digits which is exactly divisible by 24,15,36?

Answer:

The greatest number of 6 digits is 999999.

The greatest number must be divisible by L.C.M. of 24, 15 and 36

24 = 2^{3} × 3

15 = 3 × 5

36 = 2^{2} × 3^{2}

L.C.M = 2^{3} × 3^{2} × 5

= 360

To find the greatest number 999999 must be subtracted by the remainder when 999999 is divided by 360

The greatest number in 6 digits = 999999 – 279

= 999720

Question 8.

What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case?

Solution:

35 = 5 × 7

56 = 2 × 2 × 2 × 7

91 = 7 × 13

LCM of 35, 56, 91 = 5 × 7 × 2 × 2 × 2 × 13 = 3640

∴ Required number = 3647 which leaves remainder 7 in each case.

Question 9.

Find the least number that is divisible by the first ten natural numbers?

Answer:

Find the L.C.M of first 10 natural numbers

The least number is 2520

Modular Arithmetic

Two integers “a” and “b” are congruence modulo n if they differ by an integer multiple of n. That b – a = kn for some integer k. This can be written as a = b (mod n).

Euclid’s Division Lemma and Modular Arithmetic

Let m and n be integers, where m is positive. By Euclid’s division Lemma we can write n = mq + r where 0 < r < m and q is an integer.

This n = r (mod m)