Category: Class 10

Students can download Maths Chapter 3 Algebra Ex 3.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.1

Question 1.
Solve the following system of linear equations in three variables
(i) x + y + z = 5
2x – y + z = 9
x – 2y + 3z = 16
Answer:
x + y + z = 5 ….(1)
2x – y + z = 9 ….(2)
x – 2y + 3z = 16 ….(3)
by adding (1) and (2)

Substituting z = 4 (4)
3x + 2(-4) = 14
3x – 8 = 14
3x = 14 – 8
3x = 6
x = \(\frac { 6 }{ 3 } \) = 2

Substituting x = 2 and z = 4 in (1)
2 + y + 4 = 5
y + 6 = 5
y = 5 – 6
= -1
∴ The value of x = 2, y = -1 and z = 4

(ii) \(\frac { 1 }{ x } \) – \(\frac { 2 }{ y } \) + 4 = 0, \(\frac { 1 }{ y } \) – \(\frac { 1 }{ z } \) + 1 = 0, \(\frac { 2 }{ z } \) + \(\frac { 3 }{ x } \) = 14
Answer:
Let \(\frac { 1 }{ x } \) = p, \(\frac { 1 }{ y } \) = q and \(\frac { 1 }{ z } \) = r
p – 2q + 4 = 0
p – 2q = -4 ……(1)
q – r + 1 = 0
q – r = -1 ……(2)
3p + 2r = 14 ……(3)

Substituting the value of p = 2 in (1)
2 – 2q = -4
-2q = – 4 – 2
-2q = -6
q = \(\frac { 6 }{ 2 } \) = 3
Substituting the value of q = 3 in (2)
3 – r = 1
– r = – 1 – 3
r = 4

The value of x = \(\frac { 1 }{ 2 } \), y = \(\frac { 1 }{ 3 } \) and z = \(\frac { 1 }{ 4 } \)

(iii) x + 20 = \(\frac { 3y }{ 2 } \) + 10 = 2z + 5 = 110 – (y + z)
Answer:
x + 20 = \(\frac { 3y }{ 2 } \) + 10
Multiply by 2
2x + 40 = 3y + 20
2x – 3y = -40 + 20
2x – 3y = -20 ……(1)

\(\frac { 3y }{ 2 } \) + 10 = 2z + 5
Multiply by 2
3y + 20 = 4z + 10
3y – 4z = 10 – 20
3y – 4z = -10 ……(2)

2z + 5 = 110 – (y + z)
2z + 5 = 110 – y – z
y + 3z = 110 – 5
y + 3z = 105 ….(3)

Substitute the value of z = 25 in (2)
3y – 4(25) = -10
3y – 100 = – 10
3y = – 10 + 100
3y = 90
y = \(\frac { 90 }{ 3 } \) = 30
∴ The value of x = 35, y = 30 and z = 25

Substitute the value of y = 30 in (1)
2x – 3(30) = -20
2x – 90 = -20
2x = -20 + 90
2x = 70
x = \(\frac { 70 }{ 2 } \) = 35

Question 2.
Discuss the nature of solutions of the following system of equations
(i) x + 2y – z = 6, – 3x – 2y + 5z = -12 , x – 2z = 3
Answer:
x + 2y – z = 6 …..(1)
-3x – 2y + 5z = -12 …..(2)
x – 2z = 3 ……(3)
Adding (1) and (2) we get

Adding (3) and (4) we get

The above statement tells us that the system has an infinite number of solutions.

(ii) 2y + z = 3(- x + 1) ,-x + 3y – z = -4, 3x + 2y + z = –\(\frac { 1 }{ 2 } \)
2y + z = 3 (- x + 1)
2y + z = -3x + 3 ……(1)
3x + 2y + z = –\(\frac { 1 }{ 2 } \)

-x + 3y – z = – 4
x – 3y + z = 4 …..(2)

3x + 2y + z = – \(\frac { 1 }{ 2 } \)

Hence we arrive at a contradiction as 0 ≠ 7
This means that the system is inconsistent and has no solution.

(iii) \(\frac { y+z }{ 4 } \) = \(\frac { z+x }{ 3 } \) = \(\frac { x+y }{ 2 } \), x + y + z = 27
Answer:
\(\frac { y+z }{ 4 } \) = \(\frac { z+x }{ 3 } \)
3y + 3z = 4z + 4x
-4x + 3y + 3z – 4z = 0
-4x + 3y – z = 0
4x – 3y + z = 0 ………(1)

\(\frac { z+x }{ 3 } \) = \(\frac { x+y }{ 2 } \)
2z + 2x = 3x + 3y
-3x + 2x – 3y + 2z = 0
-x – 3y + 2z = 0
x + 3y – 2z = 0 ………(2)

Substituting the value of x in (5)
6 + 5z = 81
5z = 81 – 6
5z = 75
z = \(\frac { 75 }{ 5 } \) = 15
Substituting the value of x = 3
and z = 15 in (3)
3 + y + 15 = 27
y + 18 = 27
y = 27 – 18
= 9
The value of x = 3, y = 9 and z = 15
This system of equations have unique solution.

Question 3.
Vani, her father and her grand father have an average age of 53. One-half of her grand father’s age plus one-third of her father’s age plus one fourth of Vani’s age is 65. If 4 years ago Vani’s grandfather was four times as old as Vani then how old are they all now?
Answer:
Let the age of Vani be”x” years
Vani father age = “y” years
Vani grand father = “z” years
By the given first condition.
\(\frac { x+y+z }{ 3 } \) = 53
x + y + z = 159 ….(1)
By the given 2^nd Condition.
\(\frac { 1 }{ 2 } \) z + \(\frac { 1 }{ 3 } \)y + \(\frac { 1 }{ 4 } \)x = 65
Multiply by 12
6z + 4y + 3x = 780
3x + 4y + 6z = 780 ….(2)
By the given 3^rd condition
z – 4 = 4 (x – 4) ⇒ z – 4 = 4x – 16
– 4x + z = – 16 + 4
4x – z = 12 ……(3)

Vani age = 24 years
Vani’s father age = 51 years
Vani grand father age = 84 years
Substitute the value of x = 24 in (3)
4 (24) – z = 12
96 – z = 12
-z = 12 – 96
z = 84
Substitute the value of
x = 24 and z = 84 in (1)
24 + y + 84 = 159
y + 108 = 159
y = 159 – 108
= 51

Question 4.
The sum of the digits of a three-digit number is 11. If the digits are reversed, the new number is 46 more than five times the old number. If the hundreds digit plus twice the tens digit is equal to the units digit, then find the original three digit number ?
Answer:
Let the hundreds digit be x and the tens digit be ”y” and the unit digit be “z”
∴ The number is 100x + 10y + z
If the digits are reversed the new number is 100z + 10y + x
By the given first condition
x + y + z = 11 ….(1)
By the given second condition
100z + 10y + x = 5 (100x + 10y + z) + 46
= 500x + 50y + 5z + 46
x – 500x + 10y – 50y + 100z – 5z = 46
– 499x – 40y + 95z = 46
499x + 40y – 95z = -46 ….(2)
By the given third condition
x + 2y = z
x + 2y – z = 0 ….(3)

∴ The number is 137
Subtituting the value of y = 3 in (5)
2x + 3(3) = 11
2x = 11 – 9
2x = 2
x = \(\frac { 2 }{ 2 } \) = 1
Subtituting the value of x = 1, y = 3 in (1)
1 + 3 + z = 11
z = 11 – 4
= 7

Question 5.
There are 12 pieces of five, ten and twenty rupee currencies whose total value is ₹105. But when first 2 sorts are interchanged in their numbers its value will be increased by ₹20. Find the number of currencies in each sort.
Answer:
Let the number of ₹5 currencies be “x”
Let the number of ₹10 currencies be “y”
and the number of ₹20 currencies be “z”
By the given first condition
x + y + z = 12 ………(1)
By the given second condition
5x + 10y + 20z = 105
x + 2y + 4z = 21 (÷5) ……….(2)
By the given third condition
10x + 5y + 20z = 105 + 20
10x + 5y + 20z = 125
2x + y + 4z = 25 ………..(3)


Substituting the value of x = 7 in (5)
7 – y = 4 ⇒ -y = 4 – 7
-y = -3 ⇒ y = 3
Substituting the value of x = 7, y = 3 in …. (1)
7 + 3 + z = 12
z = 12 – 10 = 2
x = 7, y = 3, z = 2
Number of currencies in ₹ 5 = 7
Number of currencies in ₹ 10 = 3
Number of currencies in ₹ 20 = 2

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.6 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

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 = \(\frac { n }{ 2 } \) [2a + (n – 1) d]
S₄₀ = \(\frac { 40 }{ 2 } \) [6 + 39 × 4] = 20 [6 + 156]
= 20 × 162
S₄₀ = 3240

(ii) 102,97, 92,… up to 27 terms.
Answer:
Here a = 102, d = 97 – 102 = -5
n = 27
S_n = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S₂₇ = \(\frac { 27 }{ 2 } \) [2(102) + 26(-5)]
= \(\frac { 27 }{ 2 } \) [204 – 130]
= \(\frac { 27 }{ 2 } \) × 74
= 27 × 37 = 999
S₂₇ = 999

(iii) 6 + 13 + 20 + …. + 97
Answer:
Here a = 6, d = 13 – 6 = 7, l = 97
n = \(\frac { l-a }{ d } \) + 1
= \(\frac { 97-6 }{ 7 } \) + 1
= \(\frac { 91 }{ 7 } \) + 1 =
13 + 1 = 14
S_n = \(\frac { n }{ 2 } \) (a + l)
S_n = \(\frac { 14 }{ 2 } \) (a + l)
S_n = \(\frac { 14 }{ 2 } \) (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
\(\frac { n }{ 2 } \) [2a + (n-1) d] = 480
\(\frac { n }{ 2 } \) [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
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.
S_n = \(\frac{n}{2}\) (2a+(n – 1)d)
S₂₈ = \(\frac{28}{2}\) (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.
Answer:
Let tn be nth term of an A.P.
t_n = Sn – S_n-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
t_n = 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 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
t₁₀₄ = 125
[t_n = a + (n – 1) d]
a + 103d = 125 …..(1)
4^th 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.
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
= 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?
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 = t₂ – t₁ = 4750 – 4800 = -50
n = 10
S_n = \(\frac{n}{2}\) (2a + (n – 1)d)
S₁₀ = \(\frac{10}{2}\)  (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 = \(\frac { n }{ 2 } \) [2a + (n-1)d]
65000 = \(\frac { n }{ 2 } \) [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
∴ S_n = \(\frac{n}{2}\)  (2a + (n – 1)d)
S₃₀ = \(\frac{30}{2}\)  (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, ….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 (S₁ + S₂ + S₃ + ……. + S_m) = \(\frac { 1 }{ 2 } \) 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
S₁ + S₂ + S₃ + …… + S_m = \(\frac { n }{ 2 } \) (n + 1) + \(\frac { n }{ 2 } \) (3n + 1) + \(\frac { n }{ 2 } \) (5n + 1) + ….. + \(\frac { n }{ 2 } \) [n(2m – 1 + 1)]
= \(\frac { n }{ 2 } \) [n + 1 + 3n + 1 + 5n + 1 ……. + n (2m – 1) + m)]
= \(\frac { n }{ 2 } \) [n + 3n + 5n + ……. n(2m – 1) + m]
= \(\frac { n }{ 2 } \) [n (1 + 3 + 5 + ……(2m – 1)) + m
= \(\frac { n }{ 2 } \) [n(\(\frac { m }{ 2 } \)) (2m) + m]
= \(\frac { n }{ 2 } \) [nm² + m]
S₁ + S₂ + S₃ + ……….. + S_m = \(\frac { mn }{ 2 } \) [mn + 1]
Hint:
1 + 3 + 5 + ……. + 2m – 1
S_n = \(\frac { n }{ 2 } \) (a + l)
= \(\frac { m }{ 2 } \) (1 + 2m -1)
= \(\frac { m }{ 2 } \) (2m)

Question 12.
Find the sum

Answer:

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.7 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

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) \(\frac { 1 }{ 3 } \),\(\frac { 1 }{ 6 } \),\(\frac { 1 }{ 12 } \), ………….
(v) 1, -5, 25,-125,…
(vi) 120, 60, 30, 18,…
(vii) 16, 4, 1, \(\frac { 1 }{ 4 } \), ……….
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,
ar² = 6 × 9 = 54
The three terms are 6, 18 and 54

(ii) a = \(\sqrt { 2 }\), r = \(\sqrt { 2 }\).
Answer:
ar = \(\sqrt { 2 }\) × \(\sqrt { 2 }\) = 2,
ar² = \(\sqrt { 2 }\) × 2 = 2 \(\sqrt { 2 }\)
The three terms are \(\sqrt { 2 }\), 2 and 2\(\sqrt { 2 }\)

(iii) a = 1000, r = \(\frac { 2 }{ 5 } \)
Answer:
ar = 1000 × \(\frac { 2 }{ 5 } \) = 400,
ar² = 1000 × \(\frac { 4 }{ 25 } \) = 40 × 4 = 160
The three terms are 1000,400 and 160.

Question 3.
In a G.P. 729, 243, 81,… find t₇.
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 = \(\frac{t_{2}}{t_{1}}=\frac{t_{3}}{t_{2}}\)
\(\frac{x+12}{x+6}=\frac{x+15}{x+12}\)
(x + 12)² = (x + 6) (x + 5)
x² + 24x + 144 = x² + 6x + 15x + 90
24x – 21x = 90 – 144
3x = -54
x = \(\frac { -54 }{ 3 } \) = -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 = \(\frac { 8 }{ 4 } \) = 2
t_n = 8192

a . r^n-1 = 8192 ⇒ 4 × 2^n-1 = 8192
2^n-1 = \(\frac { 8192 }{ 4 } \) = 2048
2^n-1 = 2¹¹ ⇒ n – 1 = 11
n = 11 + 1 ⇒ n = 12
Number of terms = 12

(ii) \(\frac { 1 }{ 3 } \), \(\frac { 1 }{ 9 } \), \(\frac { 1 }{ 27 } \), ……………, \(\frac { 1 }{ 2187 } \)
Answer:
a = \(\frac { 1 }{ 3 } \) ; r = \(\frac { 1 }{ 9 } \) ÷ \(\frac { 1 }{ 3 } \) = \(\frac { 1 }{ 9 } \) × \(\frac { 3 }{ 1 } \) = \(\frac { 1 }{ 3 } \)

t_n = \(\frac { 1 }{ 2187 } \)
a. r^n-1 = \(\frac { 1 }{ 2187 } \)

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 = \(\frac { 1 }{ 2187 } \)
t₉ = 32805 [t_n = ar^n-1]
a.r⁸ = 32805 …..(1)
6^th term = 1215
a.r⁵ = 1215 …..(2)
Divide (1) by (2)
\(\frac{a r^{8}}{a r^{5}}\) = \(\frac { 32805 }{ 1215 } \) ⇒ r³ = \(\frac { 6561 }{ 243 } \)
= \(\frac { 2187 }{ 81 } \) = \(\frac { 729 }{ 27 } \) = \(\frac { 243 }{ 9 } \) = \(\frac { 81 }{ 3 } \)
r³ = 27 ⇒ r³ = 3³
r = 3
Substitute the value of r = 3 in (2)
a. 3⁵ = 1215
a × 243 = 1215
a = \(\frac { 1215 }{ 243 } \) = 5
Here a = 5, r = 3, n = 12
t₁₂ = 5 × 3^(12-1)
= 5 × 3¹¹
∴ 12^th term of a G.P. = 5 × 3¹¹

Question 7.
Find the 10th term of a G.P. whose 8^th 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 3^a, 3^b, 3^c are in G.P.
Answer:
a, b, c are in A.P.
t₂ – t₁ = t₃ – t₂
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 \(\frac { 57 }{ 2 } \). Find the three terms.
Answer:
Let the three terms of the G.P. be \(\frac { a }{ r } \), a, ar
Product of three terms = 27
\(\frac { a }{ r } \) × a × ar = 27
a³ = 27 ⇒ a³ = 3³
a = 3
Sum of the product of two terms taken at a time is \(\frac { 57 }{ 2 } \)

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 = \(\frac { 3 }{ 2 } \) (or) r = \(\frac { 2 }{ 3 } \)

∴ The three terms are 2, 3 and \(\frac { 9 }{ 2 } \) or \(\frac { 9 }{ 2 } \), 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
= \(\frac { 5 }{ 100 } \) × 60000
= ₹ 3000
Starting salary for the 2nd year = 60000 + 3000
= ₹ 63000
Year increase = 5% of 63000
= \(\frac { 5 }{ 100 } \) × 63000
= ₹ 3150
Starting salary for the 3^rd year = 63000 + 3150
= ₹ 66150
60000, 63000, 66150,…. form a G.P.
a = 60000; r = \(\frac { 63000 }{ 60000 } \) = \(\frac { 63 }{ 60 } \) = \(\frac { 21 }{ 20 } \)
t_n = an^n-1
t₅ = (60000) (\(\frac { 21 }{ 20 } \))⁴
= 60000 × \(\frac { 21 }{ 20 } \) × \(\frac { 21 }{ 20 } \) × \(\frac { 21 }{ 20 } \) × \(\frac { 21 }{ 20 } \)
= \(\frac{6 \times 21 \times 21 \times 21 \times 21}{2 \times 2 \times 2 \times 2}\)
= 72930.38
5% increase = \(\frac { 5 }{ 100 } \) × 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
= \(\frac { 5 }{ 100 } \) × 20000
= ₹ 1200
Salary for the 2nd year = ₹ 20000 + 1200
= ₹ 21200
Here a = 20,000; r = \(\frac { 21200 }{ 20000 } \) = \(\frac { 212 }{ 200 } \) = \(\frac { 106 }{ 100 } \) = \(\frac { 53 }{ 50 } \)
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
= \(\frac { 3 }{ 100 } \) × 22000
= ₹ 660
Salary for the 2nd year = ₹ 22000 + ₹ 660
= ₹ 22,660
Here a = 22000; r = \(\frac { 22660 }{ 22000 } \)
= \(\frac { 2266 }{ 2200 } \) = \(\frac { 1133 }{ 1100 } \) = \(\frac { 103 }{ 100 } \)
Salary at the end of 4th year = 22000 × (\(\frac { 103 }{ 100 } \))^4-1
= 22000 × (\(\frac { 103 }{ 100 } \))³
= 22000 × \(\frac { 103 }{ 100 } \) × \(\frac { 103 }{ 100 } \) × \(\frac { 103 }{ 100 } \)

= 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 = xr² 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

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.5 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

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…….
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) \(\frac { 1 }{ 2 } \), \(\frac { 1 }{ 3 } \), \(\frac { 1 }{ 4 } \), \(\frac { 1 }{ 5 } \), ……….
Answer:
t₂ – t₁ = \(\frac { 1 }{ 3 } \) – \(\frac { 1 }{ 2 } \) = \(\frac { 2-3 }{ 6 } \) = \(\frac { -1 }{ 6 } \)
t₃ – t₂ = \(\frac { 1 }{ 4 } \) – \(\frac { 1 }{ 3 } \) = \(\frac { 3-4 }{ 12 } \) = \(\frac { -1 }{ 12 } \)
t₂ – t₁ ≠ t₃ – t₂
The sequence is not in A.P.

(iii) 9, 13, 17, 21, 25,…
Answer:
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) \(\frac { -1 }{ 3 } \), 0, \(\frac { 1 }{ 3 } \), \(\frac { 2 }{ 3 } \)
t₂ – t₁ = 0 – (-\(\frac { 1 }{ 3 } \))
= 0 + \(\frac { 1 }{ 3 } \) = \(\frac { 1 }{ 3 } \)
t₃ – t₂ = \(\frac { 1 }{ 3 } \) – 0 = \(\frac { 1 }{ 3 } \)
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
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 = \(\frac { 3 }{ 4 } \), d = \(\frac { 1 }{ 2 } \)
Answer:
The general form of the A.P is a, a + d, a + 2d, a + 3d….
\(\frac { 3 }{ 4 } \),\(\frac { 3 }{ 4 } \) + \(\frac { 1 }{ 2 } \),\(\frac { 3 }{ 4 } \) + 2(\(\frac { 1 }{ 2 } \)), \(\frac { 3 }{ 4 } \) + 3 (\(\frac { 1 }{ 2 } \))
The A.P. \(\frac { 3 }{ 4 } \), \(\frac { 5 }{ 4 } \), \(\frac { 7 }{ 4 } \), …….

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 = 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 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
t₁₉ = -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 = t₂ – t₁ = 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 = \(\frac { 75 }{ 5 } \) =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 = \(\frac { l-a }{ d } \) + 1
= \(\frac { 183-9 }{ 6 } \) + 1
= \(\frac { 174 }{ 6 } \) + 1
= 29 + 1
= 30
middle term = 15^th term of
16^th term
t_n = 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?
Answer:
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 = \(\frac { 32 }{ 8 } \) = 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 = \(\frac { 34 }{ 2 } \) = 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
∴ 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
t_n = 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 = \(\frac{27}{3}\) = 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 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 : t₆ : t₈ = 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 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.
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 = \(\frac { 18600 }{ 600 } \) = 31
He will take 31 years to save ₹ 20,000 per month

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

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) \(\frac { 1 }{ 4 } \), \(\frac { 2 }{ 9 } \), \(\frac { 3 }{ 16 } \)
(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 \(\frac { 4 }{ 25 } \),\(\frac { 5 }{ 36 } \) and \(\frac { 6 }{ 49 } \)
[using \(\frac{n}{(n+1)^{2}}\)]

Question 2.
Find the first four terms of the sequences whose nth terms are given by
(i) a_n = n³ – 2
(ii) a_n = (-1)ⁿ⁺¹ n(n+1)
(iii) a_n = 2n² – 6
Solution:
t_n = a_n = n³ -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) a_n = (-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) a_n = 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 n^th term of the following sequences
(i) 2, 5, 10, 17, ……
Answer:
(1² + 1);(2² + 1),(3² + 1),(4² + 1)….
n^th term is n² + 1
a_n = n² + 1

(ii) 0,\(\frac { 1 }{ 2 } \),\(\frac { 2 }{ 3 } \) ……
Answer:
(\(\frac { 1-1 }{ 1 } \)), (\(\frac { 2-1 }{ 2 } \)), (\(\frac { 3-1 }{ 3 } \)) …..
n^th term is \(\frac { n-1 }{ n } \)
a_n = \(\frac { n-1 }{ n } \)

(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 = \(\frac { 5n }{ n+2 } \) ; a₆ and a₁₃
Answer:
a_n = \(\frac { 5n }{ n+2 } \)
a₆ = \(\frac { 5(6) }{ 6+2 } \) = \(\frac { 30 }{ 8 } \) = \(\frac { 15 }{ 4 } \)
a₁₃ = \(\frac { 5(13) }{ 13+2 } \) = \(\frac{5 \times 13}{15}\) = \(\frac { 13 }{ 3 } \)
a₆ = \(\frac { 15 }{ 4 } \), a₁₃ = \(\frac { 13 }{ 3 } \)

(ii) a_n = – (n² – 4); a₄ and a₁₁
Answer:
a_n = -(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 n^th term is a_n

Answer:

Question 6.
If a₁ = 1, a₂ = 1 and a_n = 2a_n-1 + a_n-2, n > 3, n ∈ N, then find the first six terms of the sequence.
Solution:
a₁ = 1, a₂ = 1, a_n = 2a_n-1 + a_n-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 ………..

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.8 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

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

Question 1.
Find the sum of first n terms of the G.P.
(i) 5, -3, \(\frac { 9 }{ 5 } \),-\(\frac { 27 }{ 25 } \), ……
(ii) 256,64,16,…….
Answer:
(i) 5,-3,\(\frac { 9 }{ 5 } \),\(\frac { 27 }{ 55 } \), ….. n terms

(ii) 256,64,16,…….
Answer:
Here a = 256, r = \(\frac { 64 }{ 256 } \) = \(\frac { 1 }{ 4 } \)

Question 2.
Find the sum of first six terms of the G.P. 5,15,45,…
Answer:
Here a = 5, r = \(\frac { 15 }{ 3 } \) = 3, n = 6

Sum of first 6 terms = 1820

Question 3.
Find the first term of the G.P. whose common ratio 5 and whose sum to first 6 terms is 46872.
Answer:
Common ratio (r) = 5
S₆ = 46872

The first term of the G.P. is 12.

Question 4.
Find the sum to infinity of (i) 9 + 3 + 1 + ….(ii) 21 + 14 + \(\frac { 28 }{ 3 } \) ……
Answer:
(i) 9 + 3 + 1 + ….
a = 9, r = \(\frac { 3 }{ 9 } \) = \(\frac { 1 }{ 3 } \)
Sum of infinity term = \(\frac { a }{ 1 – r } \) = \(\frac{9}{1-\frac{1}{3}}\)

Question 5.
If the first term of an infinite G.P. is 8 and its sum to infinity is \(\frac { 32 }{ 3 } \) then find the common ratio.
Answer:
Here a = 8, S∞ = \(\frac { 32 }{ 3 } \)
\(\frac { a }{ 1-r } \) = \(\frac { 32 }{ 3 } \)
\(\frac { 8 }{ 1-r } \) = \(\frac { 32 }{ 3 } \)
32 – 32 r = 24 ⇒ 32 r = 8
r = \(\frac { 8 }{ 32 } \) = \(\frac { 1 }{ 4 } \)
Common ration = \(\frac { 1 }{ 4 } \)

Question 6.
Find the sum to n terms of the series
(i) 0.4 + 0.44 + 0.444 + …… to n terms
Answer:


(ii) 3 + 33 + 333 + ………… to n terms
Answer:
S_n = 3 + 33 + 333 + …. to n terms
= 3[1 + 11 + 111 + …. to n terms]
= \(\frac { 3 }{ 9 } \) [9 + 99 + 999 + …. n terms]
= \(\frac { 1 }{ 3 } \) [(10 – 1) + (100 – 1) + (1000 – 1) + …… n terms]
= \(\frac { 1 }{ 3 } \) [10 + 100 + 1000 + ….. n terms – (1 + 1 + 1 ….. n terms)]

Question 7.
Find the sum of the Geometric series 3 + 6 + 12 + …….. + 1536
Answer:
3 + 6 + 12 …. +1536
a = 3, r = \(\frac { 6 }{ 3 } \) = 2
t_n = 1536


∴ Sum of the series is 3069

Question 8.
Kumar writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with the instruction that they continue the process similarly. Assuming that the process is unaltered and it costs ₹2 to mail ong letter, find the amount spent on postage when 8th set of letters is mailed.
Answer:
When kumar writes a letter to his friend.Friend writes a letter to another person.
It form a G.P
The G.P is 4, 16, 64,………
Here a = 4, r = 4
The last term is 4 (4)^8-1 = 4(4)⁷

Question 9.
Find the rational form of the number 0.123 .
Answer:
Let x = \(\overline { 0.123 } \)
= 0.123123123….
= 0.123 + 0.000123 + 000000123 + ….
This is an infinite G.P

Question 10.
If S_n = (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + ………… n terms then prove that
Answer:
S_n = (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + …….. n
Multiply by x
x S_n = x(x + y) + x(x² + xy + y²) + x(x³ + x²y + xy² + y³) + ……….. n
= x² + xy + x³ + x²y + xy² + x⁴ + x³y + x²y² + xy³ + …… n terms ……(1)
Multiply by y
yS_n = y(x + y) + y(x² + xy + y²) + y(x³ + x²y + xy² + y³) + ….. n
= xy + y² + x²y + xy² + y³ + x³y + x²y² + xy³ + y⁴ + ….. n terms
Subtract (1) and (2)
x S_n – y S_n = x² + xy + x³ + x²y + xy² + x⁴ + x³y + x²y² + xy³ + …….
– xy + y² + x²y + xy² + y³ + x³y + x²y² + xy³ + y⁴ + ……
(x – y) S_n = (x² + x³ + x⁴ + ……) – (y² + y³ + y⁴ + ……)
[ a = x²; r = x and a = y²; r = y, S_n = \(\frac{a\left(r^{n}-1\right)}{r-1}\)]

Hence it is proved.

Students can download Maths Chapter 3 Algebra Ex 3.9 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.9

Question 1.
Determine the quadratic equations, whose sum and product of roots are
(i) -9, 20
Answer:
Sum of the roots = -9 and Product of the roots = 20
The Quadratic equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (-9) x + 20 = 0 ⇒ x² + 9x + 20 = 0

(ii) \(\frac { 5 }{ 3 } \), 4
Answer:
Sum of the roots = \(\frac { 5 }{ 3 } \); Product of the roots = 4
The Quadratic equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (\(\frac { 5 }{ 3 } \)) x + 4 = 0 ⇒ x² – \(\frac { 5 }{ 3 } \) x + 4 = 0
3x² – 5x + 12 = 0

(iii) \(\frac { -3 }{ 2 } \), -1
Answer:
Sum of the roots = \(\frac { -3 }{ 2 } \); Product of the roots = -1
The Quadratic equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (-\(\frac { 3 }{ 2 } \)) x + (-1) = 0 ⇒ x² + \(\frac { 3 }{ 2 } \) x – 1 = 0
2x² + 3x – 2 = 0

(iv) – (2 – a)², (a + 5)²
Answer:
Sum of the roots = – (2 – a)²; Product of the roots = (a + 5)²
x² – (sum of the roots) x + product of the roots = 0
x² – [-(2 – a)²] x + (a + 5)² = 0
x² + (2 – a)² x + (a + 5)² = 0

Question 2.
Find the sum and product of the roots for each of the following quadratic equations
(i) x² + 3x – 28 = 0
(ii) x² + 3x = 0
(iii) 3 + \(\frac{1}{a}=\frac{10}{a^{2}}\)
(iv) 3y² – y – 4 = 0
Solution:
(i) x² – (-3)x + (-28) = 0.
Comparing this with x² – (α + β)x + αβ = 0.
(α + β) = Sum of the roots = -3
αβ = product of the roots = -28

(ii) x² + 3x = 0 = x² – (-3)x + 0 = 0
x² – (α + β)x + αβ = 0
Sum of the roots α + β = -3
Products of the roots αβ =0

(iii) 3 + \(\frac { 1 }{ a } \) = \(\frac{10}{a^{2}}\)
Answer:
Multiply by a²
3a² + a = 10
3a² + a – 10 = 0
Sum of the roots (α + β) = \(\frac { -1 }{ 3 } \)
Product of the roots (α β) = \(\frac { -10 }{ 3 } \)

(iv) 3y² – y – 4 = 0
Answer:
Sum of the roots (α + β) = \(\frac { -(-1) }{ 3 } \) = \(\frac { 1 }{ 3 } \)
Product of the roots (α β) = \(\frac { -4 }{ 3 } \)

Students can download Maths Chapter 3 Algebra Ex 3.8 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.8

Question 1.
Find the square root of the following polynomials by division method
(i) x⁴ – 12x³ + 42x² – 36x + 9
Answer:

(ii) 31 x² – 28x³ + 4x⁴ + 42x + 9
Answer:
Rearrange the order we get
4x⁴ – 28x³ + 37x² + 42x + 9

(iii) 16x⁴ + 8x² + 1
Answer:

(iv) 121 x⁴ – 198x³ – 183x² + 216x + 144
Answer:

Question 2.
Find the square root of the expression
\(\frac{x^{2}}{y^{2}}\) – \(\frac { 10x }{ y } \) + 27 – \(\frac { 10y }{ x } \) + \(\frac{y^{2}}{x^{2}}\)
Answer:

Question 3.
Find the values of a and b if the following polynomials are perfect squares.
(i) 4x⁴ – 12x³ + 37x² + bx + a
Answer:

Since it is a perfect square
b + 42 = 0
b = – 42
a – 49 = 0
a = 49
∴ The value of a = 49 and b = – 42

(ii) ax⁴ + bx³ + 361x² + 220x + 100
Answer:
Re-arrange the order we get
100 + 220x + 361x² + bx³ + ax⁴

Since it is a perfect square
b – 264 = 0
b = 264
a – 144 = 0
a = 144
∴ The value of a = 144 and b = 264

Question 4.
Find the values of m and n if the following expressions are perfect sqaures.
(i) \(\frac{1}{x^{4}}\) – \(\frac{6}{x^{3}}\) + \(\frac{13}{x^{2}}\) + \(\frac { m }{ x } \) + n
Answer:

Since it is a perfect square
\(\frac { 1 }{ x } \) (m + 12) = 0
m + 12 = 0
m = -12
n – 4 = 0
n = 4
∴ The value of m = -12 and n = 4

(ii) x⁴ – 8x³ + mx² + nx + 16
Answer:

Since it is a perfect square
m – 16 – 8 = 0
m – 24 = 0
m = 24
n + 32 = 0
n = -32
∴ The value of m = 24 and n = -32

Students can download Maths Chapter 3 Algebra Ex 3.16 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.16

1. In the

write (i) The number of elements
(ii) The order of the matrix
(iii) Write the elements a₂₂, a₂₃, a₂₄, a₃₄, a₄₃, a₄₄.
Answer:
(i) The number of elements is 16
(ii) The order of the matrix is 4 × 4
(iii) Elements corresponds to

Question 2.
If a matrix has 18 elements, what are the possible orders it can have? What if it has 6 elements?
Answer:
The possible orders of the matrix having 18 elements are

The possible orders of the matrix having 6 elements are

Question 3.
Construct a 3 × 3 matrix whose elements are given by
(i) a_ij = |i – 2j|
Answer:
a_ij = |i – 2j|
The general 3 × 3 matrices is

a₁₁ = |1 – 2(1)| = |1 – 2| = | – 1| = 1
a₁₂ = |1 – 2(2)| = |1 – 4| = | – 3| = 3
a₁₃ = |1 – 2(3)| = |1 – 6| = | – 5| = 5
a₂₁ = |2 – 2(1)| = |2 – 2| = 0 = 0
a₂₂ = |2 – 2(2)| = |2 – 4| = | – 2| = 2
a₂₃ = |2 – 2(3)| = |2 – 6| = | – 4| = 4
a₃₁ = |3 – 2(1)| = |3 – 2| = | 1 | = 1
a₃₂ = |3 – 2(2)| = |3 – 4| = | – 1 | = 1
a₃₃ = |3 – 2(3)| = |3 – 6| = | – 3 | = 3
The required matrix

(ii) a_ij = \(\frac{(i+j)^{3}}{3}\)
Answer:
a₁₁ = \(\frac{(1+1)^{3}}{3}\) = \(\frac{2^{3}}{3}\) = \(\frac { 8 }{ 3 } \)
a₁₂ = \(\frac{(1+2)^{3}}{3}\) = \(\frac { 27 }{ 3 } \) = 9
a₁₃ = \(\frac{(1+3)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₂₁ = \(\frac{(2+1)^{3}}{3}\) = \(\frac { 27 }{ 3 } \) = 9
a₂₂ = \(\frac{(2+2)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₂₃ = \(\frac{(2+3)^{3}}{3}\) = \(\frac { 125 }{ 3 } \) = \(\frac { 125 }{ 3 } \)
a₃₁ = \(\frac{(3+1)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₃₂ = \(\frac{(3+2)^{3}}{3}\) = \(\frac { 125 }{ 3 } \) = \(\frac { 125 }{ 3 } \)
a₃₃ = \(\frac{(3+3)^{3}}{3}\) = \(\frac { 216 }{ 3 } \) = 72
The required matrix

Question 4.
If  then find the tranpose of A.
Answer:

transpose of A = (A^T)

Question 5.
If then find the tranpose of – A
Answer:

Transpose of – A = (-A^T) =

Question 6.
If A = then verify (AT)T = A
Answer:

Hence it is verified

Question 7.
Find the values of x, y and z from the following equations
(i)

Answer:
Since the given matrices are equal then all the corresponding elements are equal.
y = 12, z = 3, x = 3
The value of x = 3, y = 12 and z = 3

(ii)

Answer:
x + y = 6 ……(1)
5 + z = 5
z = 5 – 5 = 0
xy = 8
y = \(\frac { 8 }{ x } \)
Substitute the value of y = \(\frac { 8 }{ x } \) in (1)
x + \(\frac { 8 }{ x } \) = 6
x² + 8 = 6x
x² – 6x + 8 = 0
(x – 4) (x – 2) = 0
∴ x – 4 = 0 or x – 2 = 0
x = 4 or x = 2
y = \(\frac { 8 }{ 4 } \) = 2 or y = \(\frac { 8 }{ 2 } \) = 4

∴ The value of x, y and z are 4, 2, 0 (or) 2, 4, 0

(iii)

Solution:
x + y + z = 9 ……….(1)
x + z = 5 ……….(2)
y + z = 7 ……….(3)

Substitute the value of y = 4 in (3)
y + z = 7
4 + z = 7
z = 7 – 4
= 3
Substitute the value of z = 3 in (2)
x + 3 = 5
x = 5 – 3
= 2
∴ The value of x = 2 , y = 4 and z = 3

Students can download Maths Chapter 3 Algebra Ex 3.7 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.7

Question 1.
Find the square root of the following.
(i) \(\frac{400 x^{4} y^{12} z^{16}}{100 x^{8} y^{4} z^{4}}\)
Answer:

(ii) \(\frac{7 x^{2}+2 \sqrt{14} x+2}{x^{2}-\frac{1}{2} x+\frac{1}{16}}\)
Answer:

(iii) \(\frac{121(a+b)^{8}(x+y)^{8}(b-c)^{8}}{81(b-c)^{4}(a-b)^{12}(b-c)^{4}}\)
Answer:

Question 2.
Find the square root of the following
(i) 4x² + 20x + 25
Answer:

(ii) 9x² – 24xy + 30xz – 40yz + 25z² + 16y²
Answer:

(iii) \(1+\frac{1}{x^{6}}+\frac{2}{x^{3}}\)
Answer:

(iv) (4x² – 9x + 2)(7x² – 13x – 2)(28x² – 3x – 1)
Answer:
4x² – 9x +2 = 4x² – 8x – x + 2
= 4x(x – 2)-1 (x – 2)
= (x – 2)(4x – 1)

7x² – 13x – 2 = 7x² – 14x + x – 2
= 7x (x – 2) + 1 (x – 2)
= (x – 2) (7x + 1)

28x² – 3x – 1 = 28x² – 7x + 4x – 1
= 7x (4x – 1) + 1 (4x – 1)
= (4x – 1) (7x + 1)

(v) (2x² + \(\frac { 17 }{ 6 } \)x + 1) (\(\frac { 3 }{ 2 } \) x² + 4x + 2) (\(\frac { 4 }{ 3 } \) x² + \(\frac { 11 }{ 3 } \) x + 2)
Answer: