## Numbers and Sequences | Aster Classes

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

Question 1.
Prove that n2 – n divisible by 2 for every positive integer n.
We know that any positive integer is of the form 2q or 2q + 1 for some integer q.
Case 1: When n = 2 q
n2 – n = (2q)2 – 2q = 4q2 – 2q
= 2q (2q – 1)
In n2 – n = 2r
2r = 2q(2q – 1)
r = q(2q + 1)
n2 – n is divisible by 2

Case 2: When n = 2q + 1
n2 – n = (2q + 1)2 – (2q + 1)
= 4q2 + 1 + 4q – 2q – 1 = 4q2 + 2q
= 2q (2q + 1)
If n2 – n = 2r
r = q (2q + 1)
∴ n2 – 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.
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.
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.
tn = a + (n – 1)d
Given tm+1 = 2 tn+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(tm+n+1)
L.H.S. = t3m+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(tm+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 12th term from the last term of the A.P -2, -4, -6,… -100.
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
tn = a + (n – 1)d
t12 = – 100 + 11 (2)
= -100 + 22
= -78
∴ The 12th 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
AP1 = a1, a1 + d, a1 + 2d,…
AP2 = a2, a2 + d, a2 + 2d,…
In AP1 we have a1 = 2
In AP2 we have a2 = 7
t10 in AP1 = a1 + 9d = 2 + 9d ………….. (1)
t10 in AP2 = a2 + 9d = 7 + 9d …………… (2)
The difference between their 10th terms
= (1) – (2) = 2 + 9d – 7 – 9d
= -5 ………….. (I)
t21 m AP1 = a1 + 20d = 2 + 20d …………. (3)
t21 in AP2 = a2 + 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?
Amount of saving in ten years = ₹ 16500
S10 = 16500, d= 100
Sn = n2 [2a + (n – 1)d]
S10 = 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–√.
2nd term of the G.P = 6–√
t2 = 6–√
[tn = a rn-1]
a.r = 6–√ ….(1)
6th term of the G.P. = 9 6–√
a. r5 = 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 1st year its value = ₹45000 – ₹6750
= ₹38250,
Again depreciation = 38250 × 15100 = 5737.50
At the end of 2nd year its value
= ₹38250 – ₹5737.50 = 32512.50
Again depreciation = 32512.50 × 15100 = 4876.88
At the end of the 3rd year its value
= 32512.50 – 4876.88 = 27635.63
∴ The value of the automobile at the 3rdyear
= ₹ 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
(1) 0, 1, 8
Hint: Let the +ve integer be 1, 2, 3, 4 …………
13 = 1 when it is divided by 9 the remainder is 1.
23 = 8 when it is divided by 9 the remainder is 8.
33 = 27 when it is divided by 9 the remainder is 0.
43 = 64 when it is divided by 9 the remainder is 1.
53 = 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
(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
(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
(4) 2520
Hint:

L.C.M. = 23 × 32 × 5 × 7
= 8 × 9 × 5 × 7
= 2520

Question 6.
74k ≡ ______ (mod 100)
(1) 1
(2) 2
(3) 3
(4) 4
(1) 1
Hint:
74k ≡______ (mod 100)
y4k ≡ y4 × 1 = 1 (mod 100)

Question 7.
Given F1 = 1 , F2 = 3 and Fn = Fn-1 + Fn-2then F5 is ………….
(1) 3
(2) 5
(3) 8
(4) 11
(4) 11
Hint:
Fn = Fn-1 + Fn-2
F3 = F2 + F1 = 3 + 1 = 4
F4 = F3 + F2 = 4 + 3 = 7
F5 = F4 + F3 = 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
(3) 7881
Hint:
t1 = 1
d = 4
tn = 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 6th term of an A.P is equal to 7 times the 7th term, then the 13th term of the A.P. is ………..
(1) 0
(2) 6
(3) 7
(4) 13
(1) 0
Hint:
6 t6 = 7 t7
6(a + 5d) = 7 (a + 6d) ⇒ 6a + 30d = 7a + 42d
30 d – 42 d = 7a – 6a ⇒ -12d = a
t13 = 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
(3) 31 m
Hint:
t16 = m
S31 = 312 (2a + 30d)
= 312 (2(a + 15d))
(∵ t16 = a + 15d)
= 31(t16) = 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
(3) 8
Here a = 1, d = 4, Sn = 120
Sn = n2[2a + (n – 1)d]
120 = n2 [2 + (n – 1)4] = n2 [2 + 4n – 4)]
= n2 [4n – 2)] = n2 × 2 (2n – 1)
120 = 2n2 – n
∴ 2n2 – n – 120 = 0 ⇒ 2n2 – 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 = 265 and B = 264 + 263 + 262 …. + 20 which of the following is true?
(1) B is 264 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
(4) A is larger than B by
A = 265
B = 264+63 + 262 + …….. + 20
= 2
= 1 + 22 + 22 + ……. + 264
a = 1, r = 2, n = 65 it is in G.P.
S65 = 1 (265 – 1) = 265 – 1
A = 265 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
(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 t1,t2,t3 … are in A.P. then the sequence t6,t12,t18 … is
(1) a Geometric Progression
(2) an Arithmetic Progression
(3) neither an Arithmetic Progression nor a Geometric Progression
(4) a constant sequence
(2) an Arithmetic Progression
Hint:
If t1, t2, t3, … is 1, 2, 3, …
If t6 = 6, t12 = 12, t18 = 18 then 6, 12, 18 … is an arithmetic progression

Question 15.
The value of (13 + 23 + 33 + ……. + 153) – (1 + 2 + 3 + …….. + 15) is …………….
(1) 14400
(2) 14200
(3) 14280
(4) 14520
(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) 62 + 72 + 82 + …….. + 212
(vi) 103 + 113 + 123 + …….. + 203
(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 = 12 + 22 + 3+ 42 + ………… + 152
15×16×316
[using n(n+1)(2n+1)6] formula
= 1240

(v) 62 + 72 + 82 + …….. + 212 = 1 + 22 + 32 + 42 + ………… + 212 – (1 + 22 + ………… + 52)
= 21×22×436 – 5×6×116
= 3311 – 55
= 3256.

(vi) 103 = 113 + 123 + …….. + 203 = 13 + 23+ 33 + ………… + 203 – (13 + 23 + 33 + …………. + 93) [Using (n(n+1)2)2 formula]
= 2102 – 452 = 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 13 + 23+ 33 + …………. + k3
1 + 2 + 3 + …. + k = 325
k(k+1)2 = 325 ……(1) = 3252 (From 1)
= 105625.

Question 3.
If 13 + 23 + 33 + ………… + K3 = 44100 then find 1 + 2 + 3 + ……. + k
13 + 23 + 33 + ………….. + k3 = 44100 k(k+1)2 = 44100−−−−−√ = 210
1 + 2 + 3 + …… + k = k(k+1)2
= 210

Question 4.
How many terms of the series 13 + 23 + 33+ …………… should be taken to get the sum 14400?
13 + 23 + 33 + ……. + n3 = 14400 n(n+1)2 = 14400−−−−−√
n(n+1)2 = 120 ⇒ n2 + n = 240 n2 + 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.
12 + 22 + 32 + …. + n2 = 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?
Area of 15 square colour papers
= 102 + 112 + 122 + …. + 242
= (12 + 22 + 32 + …. + 242) – (12 + 22 + 92)
= 24×25×496−9×10×196
= 4 × 25 × 49 – 3 × 5 × 19
= 4900 – 285
= 4615
Area can be decorated is 4615 cm2

Question 7.
Find the sum of the series (23 – 1)+(43 – 33) + (63 – 153) + …….. to
(i) n terms
(ii) 8 terms
Sum of the series = (23 – 1) + (43 – 33) + (63 – 153) + …. n terms
= 23 + 43 + 63 + …. n terms – (13 + 33 + 53+ …. n terms) …….(1)
23 + 43 + 63 + …. n = ∑(23 + 43 + 63 + ….(2n)3]
∑ 23 (13 + 23 + 33 + …. n3)
= 8 (n(n+1)2)2
= 2[n (n + 1)]2
13 + 33 + 53 + ……….(2n – 1)3 [sum of first 2n cubes – sum of first n even cubes]

Substituting (2) and (3) in (1)
Sum of the series = 2n2 (n + 1)2 – n2 (2n + 1)2 + 2n2(n + 1)2
= 4n2 (n + 1)2 – n2 (2n + 1)2
= n2 [(4(n + 1)2 – (2n + 1)2]
= n2 [4n2 + 4 + 8n – 4n2 – 1 – 4n]
= n2 [4n + 3]
= 4n3 + 3n2

(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, ………. 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 = 3Answer:a = 6, r = 3ar = 6 × 3 = 18,ar2 = 6 × 9 = 54The three terms are 6, 18 and 54(ii) a = 2–√, r = 2–√.Answer:ar = 2–√ × 2–√ = 2,ar2 = 2–√ × 2 = 2 2–√The three terms are 2–√, 2 and 22–√(iii) a = 1000, r = 25Answer:ar = 1000 × 25 = 400,ar2 = 1000 × 425 = 40 × 4 = 160The three terms are 1000,400 and 160.

Question 3.
In a G.P. 729, 243, 81,… find t7.
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)
x2 + 24x + 144 = x2 + 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?
Here a = 4; r = 84 = 2
tn = 8192

a . rn-1 = 8192 ⇒ 4 × 2n-1 = 8192
2n-1 = 81924 = 2048
2n-1 = 211 ⇒ n – 1 = 11
n = 11 + 1 ⇒ n = 12
Number of terms = 12

(ii) 13, 19, 127, ……………, 12187
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 9th term is 32805 and 6th term is 1215. Find the 12th term.
Given, 9th term = 32805
a. rn-1 = 12187
t9 = 32805 [tn = arn-1]
a.r8 = 32805 …..(1)
6th term = 1215
a.r5 = 1215 …..(2)
Divide (1) by (2)
ar8ar5 = 328051215 ⇒ r3 = 6561243
= 218781 = 72927 = 2439 = 813
r3 = 27 ⇒ r3 = 33
r = 3
Substitute the value of r = 3 in (2)
a. 35 = 1215
a × 243 = 1215
a = 1215243 = 5
Here a = 5, r = 3, n = 12
t12 = 5 × 3(12-1)
= 5 × 311
∴ 12th term of a G.P. = 5 × 311

Question 7.
Find the 10th term of a G.P. whose 8th term is 768 and the common ratio is 2.
Solution:
t8 = 768 = ar7
r = 2
t10 = ar9 = ar7 × r × r
= 768 × 2 × 2 = 3072

Question 8.
If a, b, c are in A.P. then show that 3a, 3b, 3c are in G.P.
a, b, c are in A.P.
t2 – t1 = t3 – t2
b – a = c – b
2b = a + c …..(1)
3a, 3b, 3c are in G.P.

From (1) and (2) we get
3a, 3b, 3c 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.
Let the three terms of the G.P. be ar, a, ar
Product of three terms = 27
ar × a × ar = 27
a3 = 27 ⇒ a3 = 33
a = 3
Sum of the product of two terms taken at a time is 572 6r2 – 13r + 6 = 0
6r2 – 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?
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 3rd year = 63000 + 3150
= ₹ 66150
60000, 63000, 66150,…. form a G.P.
a = 60000; r = 6300060000 = 6360 = 2120
tn = ann-1
t5 = (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 4th year with respect to the offers A and B?
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
tn = arn-1

Salary at the end of 4th 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
4th year Salary for A = ₹ 23820 and 4th 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 xb-c × yc-a × za-b = 1
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 = xr2 respective ……(2)
L.H.S = xb-c × yc-a × za-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, ……….

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
a = 6, r = 3
ar = 6 × 3 = 18,
ar2 = 6 × 9 = 54
The three terms are 6, 18 and 54

(ii) a = 2–√, r = 2–√.
ar = 2–√ × 2–√ = 2,
ar2 = 2–√ × 2 = 2 2–√
The three terms are 2–√, 2 and 22–√

(iii) a = 1000, r = 25
ar = 1000 × 25 = 400,
ar2 = 1000 × 425 = 40 × 4 = 160
The three terms are 1000,400 and 160.

Question 3.
In a G.P. 729, 243, 81,… find t7.
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)
x2 + 24x + 144 = x2 + 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?
Here a = 4; r = 84 = 2
tn = 8192

a . rn-1 = 8192 ⇒ 4 × 2n-1 = 8192
2n-1 = 81924 = 2048
2n-1 = 211 ⇒ n – 1 = 11
n = 11 + 1 ⇒ n = 12
Number of terms = 12

(ii) 13, 19, 127, ……………, 12187
a = 13 ; r = 19 ÷ 13 = 19 × 31 = 1

tn = 12187
a. rn-1 = 12187

n – 1 = 6 ⇒ n = 6 + 1 = 7
Number of terms = 7

Question 6.
In a G.P. the 9th term is 32805 and 6th term is 1215. Find the 12th term.
Given, 9th term = 32805
a. rn-1 = 12187
t9 = 32805 [tn = arn-1]
a.r8 = 32805 …..(1)
6th term = 1215
a.r5 = 1215 …..(2)
Divide (1) by (2)
ar8ar5 = 328051215 ⇒ r3 = 6561243
= 218781 = 72927 = 2439 = 813
r3 = 27 ⇒ r3 = 33
r = 3
Substitute the value of r = 3 in (2)
a. 35 = 1215
a × 243 = 1215
a = 1215243 = 5
Here a = 5, r = 3, n = 12
t12 = 5 × 3(12-1)
= 5 × 311
∴ 12th term of a G.P. = 5 × 311

Question 7.
Find the 10th term of a G.P. whose 8th term is 768 and the common ratio is 2.
Solution:
t8 = 768 = ar7
r = 2
t10 = ar9 = ar7 × r × r
= 768 × 2 × 2 = 3072

Question 8.
If a, b, c are in A.P. then show that 3a, 3b, 3c are in G.P.
a, b, c are in A.P.
t2 – t1 = t3 – t2
b – a = c – b
2b = a + c …..(1)
3a, 3b, 3c are in G.P.

From (1) and (2) we get
3a, 3b, 3c 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.

Let the three terms of the G.P. be ar, a, ar
Product of three terms = 27
ar × a × ar = 27
a3 = 27 ⇒ a3 = 33
a = 3
Sum of the product of two terms taken at a time is 572 6r2 – 13r + 6 = 0
6r2 – 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?

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 3rd year = 63000 + 3150
= ₹ 66150
60000, 63000, 66150,…. form a G.P.
a = 60000; r = 6300060000 = 6360 = 2120
tn = ann-1
t5 = (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 4th year with respect to the offers A and B?

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
tn = arn-1

Salary at the end of 4th 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
4th year Salary for A = ₹ 23820 and 4th 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 xb-c × yc-a × za-b = 1
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 = xr2 respective ……(2)
L.H.S = xb-c × yc-a × za-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.
3,7,11,… up to 40 terms
First term (a) = 3
Common difference (d) = 7 – 3 = 4
Number of terms (n) = 40
Sn = n2 [2a + (n – 1) d]
S40 = 402 [6 + 39 × 4] = 20 [6 + 156]
= 20 × 162
S40 = 3240

(ii) 102,97, 92,… up to 27 terms.
Here a = 102, d = 97 – 102 = -5
n = 27
Sn = n2 [2a + (n – 1)d]
S27 = 272 [2(102) + 26(-5)]
= 272 [204 – 130]
= 272 × 74
= 27 × 37 = 999
S27 = 999

(iii) 6 + 13 + 20 + …. + 97
Here a = 6, d = 13 – 6 = 7, l = 97
= 97−67 + 1
= 917 + 1 =
13 + 1 = 14
Sn = n2 (a + l)
Sn = 142 (a + l)
Sn = 142 (6 + 97)
= 7 × 103
Sn = 721

Question 2.
How many consecutive odd integers beginning with 5 will sum to 480?
First term (a) = 5
Common difference (d) = 2
(consecutive odd integer)
Sn = 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 nth term is 4n – 3.
Solution:
n = 28
tn = 4n – 3
t1 = 4 × 1 – 3 = 1
t2 = 4 × 2 – 3 = 5
t28 = 4 × 28 – 3
= 112 – 3 = 109
∴ a = 1, d = t2 – t1 = 5 – 1 = 4
l = 109.
Sn = n2 (2a+(n – 1)d)
S28 = 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 2n2 – 3n . Show that the series is an A.P.
Let tn be nth term of an A.P.
tn = Sn – Sn-1
= 2n2 – 3n – [2(n – 1)2 – 3(n – 1)]
= 2n2 – 3n – [2(n2 – 2n + 1) – 3n + 3]
= 2n2 – 3n – [2n2 – 4n + 2 – 3n + 3]
= 2n2 – 3n – [2n2 – 7n + 5]
= 2n2 – 3n – 2n2 + 7n – 5
tn = 4n – 5
t1 = 4(1) – 5 = 4 – 5 = -1
t2 = 4(2) -5 = 8 – 5 = 3
t3 = 4(3) – 5 = 12 – 5 = 7
t4 = 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 4th term of an A.P are 125 and 0. Find the sum of first 35 terms?
104th term of an A.P = 125
t104 = 125
[tn = a + (n – 1) d]
a + 103d = 125 …..(1)
4th term = 0
t4 = 0
a + 3d = 0 …..(2) Sum of 35 terms = 612.5

Question 6.
Find the sum of ail odd positive integers less than 450.
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
= n2
= 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?
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 = Sn1 – Sn2
= 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 = t2 – t1 = 4750 – 4800 = -50
n = 10
Sn = n2 (2a + (n – 1)d)
S10 = 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?
(i) Total loan amount = ₹ 65,000
Sn = 65,000
First month payment (a) = 400
Every month increasing ₹ 300
d = 300
Sn = n2 [2a + (n-1)d]
65000 = n2 [2(400) + (n – 1)300]
130000 = n [800 + 300n – 300]
= n [500 + 300n]
13000 = 500n + 300n2
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
∴ Sn = n2  (2a + (n – 1)d)
S30 = 302  (2 × 100 + 29 × -2)
= 15(200 – 58)
= 15 × 142
= 2130
t30 = 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 S1, S2 , S3, ….Sm 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 (S1 + S2 + S3 + ……. + Sm) = 12 mn(mn + 1)
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
S1 + S2 + S3 + …… + Sm = 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 [nm2 + m]
S1 + S2 + S3 + ……….. + Sm = mn2 [mn + 1]
Hint:
1 + 3 + 5 + ……. + 2m – 1
Sn = n2 (a + l)
= m2 (1 + 2m -1)
= m2 (2m)

Question 12.
Find the sum 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,…
a – 3, a – 5, a – 7…….
t2 – t1 = a – 5 – (a – 3)
= a – 5 – a + 3
= -2
t3 – t2 = a – 7 – (a – 5)
= a – 7 – a + 5
= -2
t2 – t1 = t3 – t2
(common difference is same)
The sequence is in A.P.

(ii) 12, 13, 14, 15, ……….
t2 – t1 = 13 – 12 = 2−36 = −16
t3 – t2 = 14 – 13 = 3−412 = −112
t2 – t1 ≠ t3 – t2
The sequence is not in A.P.

(iii) 9, 13, 17, 21, 25,…
t2 – t1 = 13 – 9 = 4
t3 – t2 = 17 – 13 = 4
t4 – t3 = 21 – 17 = 4
t5 – t4 = 25 – 21 = 4
Common difference are equal
∴ The sequence is in A.P.

(iv) −13, 0, 13, 23
t2 – t1 = 0 – (-13)
= 0 + 13 = 13
t3 – t2 = 13 – 0 = 13
t2 – t1 = t3 – t2
The sequence is in A.P.

(v) 1,-1, 1,-1, 1, -1, …
t2 – t1 = -1 – 1 = -2
t3 – t2 = 1 – (-1) = 1 + 1 = 2
t4 – t3 = -1-(1) = – 1 – 1 = – 2
t5 – t4 = 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
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
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
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 nth terms are given below
(i) tn = -3 + 2n
(ii) tn = 4 – 7n
Solution:
(i) a = t1 = -3 + 2(1) = -3 + 2 = -1
d = t2 – t1
Here t2 = -3 + 2(2) = -3 + 4 = 1
∴ d = t2 – t1 = 1 – (-1) = 2
(ii) a = t1 = 4 – 7(1) = 4 – 7 = -3
d = t2 – t1
Here t2 = 4 – 7(2) = 4 – 14 – 10
∴ d = t2 – t1 = 10 – (-3) = -7

Question 4.
Find the 19th term of an A.P. -11, -15, -19,…
First term (a) = -11
Common difference (d) = -15 -(-11)
= -15 + 11 = -4
n = 19
tn = a + (n – 1) d
tn = -11 + 18(-4)
= -11 – 72
t19 = -83
19th 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
tn = -54
a = 16, d = t2 – t1 = 11 – 16 = -5
∴ tn = 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.
First term (a) = 9
Last term (l) = 183
Common difference (d) = 15 – 9 = 6
= 183−96 + 1
= 1746 + 1
= 29 + 1
= 30
middle term = 15th term of
16th term
tn = a + (n – 1)d
t15 = 9 + 14(6)
= 9 + 84 = 93
t16 = 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
9t9 = 15t15
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
⇒ 6t24 = 0. Hence it is proved.

Question 8.
If 3 + k, 18 – k, 5k + 1 are in A.P. then find k?
3 + k, 18 – k, 5k + 1 are in AP
∴ t2 – t1 = t3 – t2 (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 = t2 – t1 = 10 – x ………….. (1)
= t3 – t2 = y – 10 ………….. (2)
= t4 – t3 = 24 – y …………. (3)
= t5 – t4 = 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?
Number of seats in the first row
(a) = 20
∴ t1 = 20
Number of seats in the second row
(t2) = 20 + 2
= 22
Number of seats in the third row
(t3) = 22 + 2
= 24
Here a = 20 ; d = 2
Number of rows
(n) = 30
tn = a + (n – 1)d
t30 = 20 + 29(2)
= 20 + 58
t30 = 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(a2 – d2) = 288
⇒ 9(9 – d2) = 288
⇒ 9(81 – d2) = 288
81 – d2 = 32
-d2 = 32 – 81
d2 = 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 6th and 8th term of an A.P is 7:9. Find the ratio of 9th term to 13th term.
Given : t6 : t8 = 7 : 9 (using tn = 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 t9 : t13
t9 : t13 = a + 8d : a + 12d
= 2d + 8d : 2d + 12d
= 10d : 14d
= 5 : 7
∴ t9 : t13 = 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.
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)

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.
Tabulate the given table Monthly savings form an A.P.
2000, 2600, 3200 …..
a = 2000; d = 2600 – 2000 = 600
Given tn = 20,000
tn = 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
(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) an = n3 – 2
(ii) an = (-1)n+1 n(n+1)
(iii) an = 2n2 – 6
Solution:
t= a= n-2
(i) a1 = 13 – 2 = 1 – 2 – 1
a2 = 23 – 2 = 8 – 2 = 6
a3 = 33 – 2 = 27 – 2 = 25
a4 = 43 – 2 = 64 – 2 = 62
∴ The first four terms are -1, 6, 25, 62, ……….

(ii) an = (-1)n+1 n(n + 1)
a1 = (-1)1+1 (1) (1 +1)
= (-1)2 (1) (2) = 2
a2 = (-1)2+1 (2) (2 + 1)
= (-1)3 (2) (3)= -6
a3 = (-1)3+1 (3) (3 + 1)
= (-1)4 (3) (4) = 12
a4 = (-1)4+1 (4) (4 + 1)
= (-1)5 (4) (5) = -20
∴ The first four terms are 2, -6, 12, -20,…

(iii) an = 2n2 – 6
a1 = 2(1)2 – 6 = 2 – 6 = -4
a2 = 2(2)2 – 6 = 8 – 6 = 2
a3 = 2(3)2 – 6 = 18 – 6 = 12
a4 = 2(4)2 – 6 = 32 – 6 = 26
∴ The first four terms are -4, 2, 12, 26, …

Question 3.
Find the nth term of the following sequences
(i) 2, 5, 10, 17, ……
(12 + 1);(22 + 1),(32 + 1),(42 + 1)….
nth term is n2 + 1
an = n2 + 1

(ii) 0,12,23 ……
(1−11), (2−12), (3−13) …..
nth term is n−1n
an = n−1n

(iii) 3,8,13,18,…….
[5(1) -2], [5(2) – 2], [5(3) – 2], [5(4) – 2] ….
The nth term is 5n – 2
an = 5n – 2

Question 4.
Find the indicated terms of the sequences whose nth terms are given by

(i) an = 5nn+2 ; a6 and a13
an = 5nn+2
a6 = 5(6)6+2 = 308 = 154
a13 = 5(13)13+2 = 5×1315 = 133
a6 = 154, a13 = 133

(ii) an = – (n2 – 4); a4 and a11
an = -(n2 – 4)
a4 = -(42 – 4)
= – (16 – 4)
= -12
a11 = -(112 – 4)
= – (121 – 4)
= – 117
a4 = -12 and a11 = -117

Question 5.
Find a8 and a15 whose nth term is an

Question 6.
If a1 = 1, a2 = 1 and an = 2an-1 + an-2, n > 3, n ∈ N, then find the first six terms of the sequence.
Solution:
a1 = 1, a2 = 1, an = 2an-1 + an-2
a3 = 2a(3-1) + a(3-2)
= 2a2 + a1
= 2 × 1 + 1 = 3
a4 = 2a(4-1) + a(4-2)
= 2a3 + a2
= 2 × 3 + 1 = 7
a5 = 2a(5-1) + a(5-2)
= 2a4 + a3
= 2 × 7 + 3 = 17
a6 = 2a(6-1) + a(6-2)
= 2a5 + a4
= 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)
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)
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.?
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?
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 2n + 6 × 9n is always divisible by 7 for any positive integer n.
Solution:
21 + 6 × 91 = 2 + 54 = 56 is divisible by 7
When n = k,
2k + 6 × 9k = 7 m [where m is a scalar]
⇒ 6 × 9k = 7 m – 2k …………. (1)
Let us prove for n = k + 1
Consider 2k+1 + 6 × 9k+1 = 2k+1 + 6 × 9k × 9
= 2k+1 + (7m – 2k)9 (using (1))
= 2k+1 + 63m – 9.2k = 63m + 2k.21 – 9.2k
= 63m – 2k (9 – 2) = 63m – 7.2k
= 7 (9m – 2k) which is divisible by 7
∴ 2n + 6 × 9n is divisible by 7 for any positive integer n

Question 9.
Find the remainder when 281 is divided by 17?
281 ≡ x(mod 17)
240 × 240 × 21 ≡ x(mod 17)
(24)10 × (24)10 × 21 ≡ x(mod 17)
(16)10 × (16)10 × 21 ≡ x(mod 17)
(162)5 × (162)5 × 21 ≡ x(mod 17)
= 1 × 1 × 2 (mod 17)
[(16)2 = 256 = 1 (mod 17)]
= 2 (mod 17)
281 = 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?
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?
4n = (22)n = 22n
= 2n × 2n
2 is a factor of 4n
∴ 4n is always even.

Question 2.
If m, n are natural numbers, for what values of m, does 2n × 5n ends in 5?
Solution:
2n × 5m
2n is always even for all values of n.
5m is always odd and ends with 5 for all values of m.
But 2n × 5m is always even and ends in 0.
∴ 2n × 5m cannot end with the digit 5 for any values of m. No value of m will satisfy 2n × 5m ends in 5.

Question 3.
Find the H.C.F. of 252525 and 363636.
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 = 2a × 3b then find a and b?
Using factor tree method factorise 13824 13824 = 29 × 33
Given 13824 = 2a × 3b
Compare we get a = 9 and b = 3

Aliter:
13824 = 29 × 33
Compare with
13824 = 2a × 3b
The value of a = 9 b = 3

Question 5.
If p1x1 × p2x2 × p3x3 × p4x4 = 113400 where p1 p2, p3, p4 are primes in ascending order and x1, x2, x3, x4, are integers, find the value of p1,p2,p3,p4 and x1,x2,x3,x4.
Given 113400 = p1x1 × p2x2 × p3x3 × p4x4
Using tree method factorize 113400

113400 = 23 × 34 × 52 × 7
compare with
113400 = p1x1 × p2x2 × p3x3 × p4x4
P1 = 2, x1 = 3
P2 = 3, x2 = 4
P3 = 5, x3 = 2
P4 = 7, x4 = 1

Question 6.
Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of Arithmetic.
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?
The greatest number of 6 digits is 999999.
The greatest number must be divisible by L.C.M. of 24, 15 and 36

24 = 23 × 3
15 = 3 × 5
36 = 22 × 32
L.C.M = 23 × 32 × 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?
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)

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